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Projectile Motion Notes - PowerPoint PPT Presentation

Projectile Motion Notes. Vertical Projectile Motion. Vertical Projectile Motion. Case 1 Case 2 Case 3 Case 4 . Object launched vertically and returns to launch height. Object launched vertically and returns to a different height. u. u. Object projected downwards.

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Vertical Projectile Motion

Case 1 Case 2 Case 3 Case 4

Object launched vertically and returns to launch height

Object launched vertically and returns to a different height

u

u

Object projected downwards

Object dropped from rest

u

Case 1: Vertical Projectile Motion

Set positive direction as down

+

Use constant acceleration equations with u = 0 a = 10ms-1

If time not involved can use energy approach

Ug Ek

mgh = ½mv2

gh= ½v2

2gh= v2

Three points

Case 2: Vertical Projectile Motion

Set positive direction as down

+

Use constant acceleration equations with a = 10ms-1

u

If time not involved can use energy approach

Eki+ Ug Ekf

½mu2 + mgh = ½mv2

½u2+ gh= ½v2

u2+ 2gh= v2

Three points

Case 3: Vertical Projectile Motion

If time not involved can use energy approach

Eki Ekf + Ug

½mu2 = ½mv2 + mgh

½u2 = ½v2 + gh

u2 = v2 + 2gh

v = 0

+

Speeds up are the same as down at each height

Use constaccel equations with a = -10ms-1

u

v

v

Time up = time down

or

Total time = 2 × time up

Set positive direction as up

Six points

Case 4: Vertical Projectile Motion

If time not involved can use energy approach

Eki Ekf + Ug

½mu2 = ½mv2 + mgh

v = 0

+

Speeds up are the same as down at each height

Use constaccel equations with a = -10ms-1

u

Displacements below launch height will be negative

Set positive direction as up

Six points

Case 4: Vertical Projectile Motion

If time not involved can use energy approach

Eki Ekf + Ug

½mu2 = ½mv2 + mgh

v = 0

+

Speeds up are the same as down at each height

Use constaccel equations with a = -10ms-1

u

Displacements below launch height will be negative

Set positive direction as up

Six points

Vertical Projectile Motion

Worked Examples

Vertical Projectile MotionExample 1

A stone is dropped from an 8.0m tower and falls to the ground.

(a)How long will it take the stone to drop to the ground?

t = ? u = 0 x = 8.0m a = 10ms-2

x = ut + ½ at2

8 = ½ × 10 × t2

8 = 5 × t2

1.6 = t2

1.26491 = t

t 1.3 s

+

• In Maths when you determine the square root of a number you should always give the result as a  value. This means there are two answers

• In Physics we normally just write out the magnitude of the square root because the negative value is eithernot relevant orit represents a direction which is already obvious in the problem. In this case a negative time is not plausible.

Vertical Projectile MotionExample 1

A stone is dropped from an 8.0m tower and falls to the ground.

(b)What speed will the stone hit the ground?

v = ? u = 0 x = 8.0m a = 10ms-2

v2 = u2 + 2ax

v2= 2× 10 × 8

v2 = 160

v = 12.6491

v 13 ms-1

+

Alternative – Energy Approach

Ug Ek

mgh = ½mv2

gh = ½v2

2gh= v2

2× 10 × 8 = v2

160 = v2

12.6491 = v

v = 13ms-1

• In In Physics we normally just write out the magnitude of the square root because the negative value is eithernot relevant orit represents a direction which is already obvious in the problem. In this case a negative value means the stone is moving upwards which is not plausible.

Vertical Projectile MotionExample 2

A stone is dropped down a well and takes 2.5 seconds to hit the water. How deep is the well?

x = ? u = 0 t= 2.5s a = 10ms-2

x = ut + ½ at2

x= ½ × 10 × 2.52

x= 31.25

x 31 s

+

Vertical Projectile MotionExample 3

A stone is thrown down at 2.0ms-1 from a 3.0m tower and falls to the ground. What speed did the ground hit the ground?

v = ? u = 0 x = 10m a = 10ms-2

v2 = u2 + 2ax

v2= 22+ 2 × 10 × 3

v2 = 64

v = 8 ms-1

+

Alternative – Energy Approach

Eki+ Ug Ekf

½mu2 + mgh = ½mv2

½u2 + gh = ½v2

u2 + 2gh = v2

22+ 2 × 10 × 3 = v2

64 = v2

8 = v

v = 8ms-1

2.0ms-1

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(a) How high will the cannon ball reach?

x = ? u = 30ms-1v= 0 a = –10ms-2

v2= u2 + 2ax

0= 302+ 2 ×– 10 × x

0 = 900 + – 20 × x

– 900 = – 20 × x

45 = x

x= 45m

+

30ms-1

x

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(a) How high will the cannon ball reach?

x = ? u = 30ms-1v= 0 a = –10ms-2

v2= u2 + 2ax

0= 302+ 2 ×– 10 × x

0 = 900 + – 20 × x

– 900 = – 20 × x

45 = x

x= 45m

+

30ms-1

Alternative – Energy Approach

EkUg

½mv2 = mgh

½v2 = gh

v2 = 2gh

302 = 2 × 10 × h

900 = 20h

45 = h

h = 45m

x

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(b) What will be the flight time of the cannon ball?

Find time to top of the path

t = ? u= 30ms-1v= 0 a = –10ms-2

v= u+ at

0= 30 + – 10 × t

– 30 = – 10 × t

3 = t

Find total time

Total time = 2× 3

= 6 seconds

+

30ms-1

t

t

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(b) What will be the flight time of the cannon ball?

Find time to top of the path

t = ? u= 30ms-1v= 0 a = –10ms-1

v= u+ at

0= 30 + – 10 × t

– 30 = – 10 × t

3 = t

Find total time

Total time = 2× 3

= 6 seconds

Alternative 1

t = ? u = 30ms-1x = 0 a = – 10ms-2

x = ut + ½at2

0= 30t+ ½ × – 10 × t2

0= 30t + – 5 × t2

0= 5t (6 – t)

t= 0, 6

So flight time = 6 seconds

+

30ms-1

t

t

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(b) What will be the flight time of the cannon ball?

Find time to top of the path

t = ? u= 30ms-1v= 0 a = –10ms-1

v= u+ at

0= 30 + – 10 × t

– 30 = – 10 × t

3 = t

Find total time

Total time = 2× 3

= 6 seconds

Alternative 2

t = ? u = 30ms-1v = – 30ms-1 a = – 10ms-2

v= u + at

– 30= 30 + – 10 × t

– 60 = – 10 × t

6 = t

So flight time = 6 seconds

+

30ms-1

t

t

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(c) What will be the speed at a height of 30m?

t = ? u= 30ms-1x= 30ma = –10ms-2

v2 = u2 + 2ax

v2= 302 + 2 × – 10 × 30

v2 = 300

v=  17.321 ms-1

So the speed will be 17ms-1at 30m on the way up and 17ms-1at 30m the way down

+

30ms-1

v

v

30m

Vertical Projectile MotionExample 4

A cannon ball is fired upwards from the ground at a speed of 30ms-1.

(c) What will be the speed at a height of 30m?

t = ? u= 30ms-1x= 30ma = –10ms-2

v2 = u2 + 2ax

v2= 302 + 2 × – 10 × 30

v2 = 300

v=  17.321 ms-1

So the speed will be 17ms-1at 30m on the way up and 17ms-1at 30m the way down

Alternative Using energy approach

v = ? u = 30ms-1h = 30m g = 10ms-1

Eki Ekf+ Ug

½mu2= ½mv2 + mgh

½u2= ½v2+ gh

u2= v2+ 2gh

302= v2+ 2×10×30

900 = v2 + 600

300 = v2

17.3205 = v

v  17ms-1

+

30ms-1

v

v

30m

Vertical Projectile MotionExample 5

If a cannon ball fired from 4.0m above the ground is in the air for 8.0 seconds, what is the launch speed of the cannon ball?

u = ? x= – 4.0m t = 8.0s a = –10ms-2

x = ut + ½ at2

– 4 = u × 8+ ½ × – 10 × 82

– 4 = 8u – 320

316 = 8u

39.5 = u

u 40 ms-1

+

4.0m

Vertical Projectile Motion

Exam Questions

2006 ExamQ6 & 7

When you have F, time and no distance you can quite often use formulae from the momentum/impulse string

I = Ft = p = pf – pi = m(v– u)

to solve the problem

A rocket of mass 0.50 kg is set on the ground, pointing vertically up. When ignited, the gunpowder burns for a period of 1.5 s, and provides a constant force of 22 N. The mass of the gunpowder is very small compared to the mass of the rocket, and can be ignored.

After 1.5 s, what is the height of the rocket above the ground?

+

22N

v= ? t= 1.5su = 0 a= = = 34ms-2

v = u + at

v= 0 + 34 × 1.5

v = 51 ms-1

alternative using momentum/impulse approach

v = ? t = 1.5su = 0 Fnet = 17N

Ft= m(v – u)

17 × 1.5 = 0.5v

25.5 = 0.5v

51 ms-1 = v

W = mg

= 0.5× 10

= 5N

Horizontally Launched Projectile Motion

Horizontally Launched Projectile Motion

Case 1 Case 2

Horizontally launched object drops to ground level

Ex1 – object launched of building/cliff

Ex2 – object falling off a moving vehicle

Ex3 – object dropped from an aircraft

u

u

u

Horizontally Launched Projectile Motion

Case 1 Case 2

Horizontally launched object drops to another height

Ex – object launched of building to the top of another building

u

u

Case 2: Horizontally Launched Projectile Motion

Reference

axes

The motion in the x & y directions are independent

vyis accelerated by gvxis constant = u

In y direction uy = 0 a = 10ms-1

and use constaccelequns

x

In x dirnvx= u

and use

u

vx

vx

vx

If not enough information in one direction get time from the other direction

d

vy

vy

t

v

y

vy

If time not involved

Eki+ Ug Ekf

½mu2 + mgh = ½mv2

u2+ 2gh= v2

Six points

Case 2: Horizontally Launched Projectile Motion

On landing

thproj=

x

Rhproj = u

u

vx

vx

vx

v =

What do and u represent?

vx & vy

on landing

vy

vy

y

vy

•  = tan-1

Four landing formulae

Six points

v

Case 2: Horizontally Launched Projectile Motion

If you treat the top of the second building as a virtual ground level all the previous five points including formulae hold for this situation

h2

h = h1 – h2

h1

u

Virtual ground level

Four landing formulae

Six points

Horizontal Projectile Motion

Worked Examples

360kmh-1

= 100ms-1

An aeroplane travelling at 360kmh-1 drops a food package when it passes over a hiker at a height of 300m.

(a)How far from the hiker will the food land?

In x direction

d = ? v = ux= 100ms-1t = ?

d = vt

d = 100 × 7.74597

d = 774.597

d  775m

x

d = ?

Find t from the y direction

t = ? u= 0 x=8.0m a= 10ms-1

x= ut + ½ at2

300= ½ × 10 × t2

300 = 5 × t2

60 = t2

7.74597 = t

7.74597s

y

360kmh-1

= 100ms-1

An aeroplane travelling at 360kmh-1 dropped a food package when it passes over a hiker at a height of 300m.

(a)How far from the hiker will the food land?

In x direction

d = ? v = ux= 100ms-1t = ?

d = vt

d = 100 × 7.74597

d = 774.597

d  775m

x

d = ?

Alternative using derived formulae

Rhproj = ? u = 100ms-1h = 300m g = 10ms-1

Rhproj = u

Rhproj=

Rhproj= 774.597

Rhproj 775m

Find t from the y direction

t = ? u= 0 x=8.0m a= 10ms-1

x= ut + ½ at2

300= ½ × 10 × t2

300 = 5 × t2

60 = t2

7.74597 = t

7.74597s

y

360kmh-1

= 100ms-1

An aeroplane travelling at 360kmh-1 dropped a food package when it passes over a hiker at a height of 300m.

(b)What will be the velocity of the package when it hits the ground?

In y direction

vy = ? uy= 0 x=300m a = 10ms-1

v2 = u2 + 2ax

v2 = 2 × 10 × 300

v2=6000

v = 77.4597

x

Find final velocity

h2 = a2 + b2

v2= 1002+ 77.45072

v2= 16000

v= 126.49

v 126 ms-1

v = ?

100ms-1

77.4597ms-1

v

tan  =

tan  =

 = tan-1

 = 37.761o

  38o

y

38o

126 ms-1

360kmh-1

= 100ms-1

An aeroplane travelling at 360kmh-1 dropped a food package when it passes over a hiker at a height of 300m.

(b)What will be the velocity of the package when it hits the ground?

In y direction

vy = ? uy= 0 x=300m a = 10ms-1

v2 = u2 + 2ax

v2 = 2 × 10 × 300

v2=6000

v = 77.4597

x

Find final velocity

h2 = a2 + b2

v2= 1002+ 77.45072

v2= 16000

v= 126.49

v 126 ms-1

v = ?

100ms-1

38o

Alternative using derived formulae

v = ? u = 100ms-1h = 300m g = 10ms-1

v =

v =

v = 126.491

v  126ms-1

77.4597ms-1

38o

v

126 ms-1

126 ms-1

y

•  = tan-1

•  = tan-1

•  = 37.761o

•  38o

tan  =

tan  =

 = tan-1

 = 37.761o

  38o

360kmh-1

= 100ms-1

An aeroplane travelling at 360kmh-1 dropped a food package when it passes over a hiker at a height of 300m.

(b)What will be the velocity of the package when it hits the ground?

In y direction

vy = ? uy= 0 x=300m a = 10ms-1

v2 = u2 + 2ax

v2 = 2 × 10 × 300

v2=6000

v = 77.4597

x

v = ?

Find final velocity

h2 = a2 + b2

v2= 1002+ 77.45072

v2= 16000

v= 126.49

v 126 ms-1

Alternative if just asked for speed (energy approach)

v = ? u = 100ms-1h = 300m g = 10ms-1

Eki+ Ug Ekf

½mu2 + mgh = ½mv2

½u2+ gh= ½v2

u2 + 2gh = v2

1002+ 2×10×300 = v2

16000 = v2

126.491 = v

v  126ms-1

100ms-1

38o

v

126 ms-1

y

tan  =

tan  =

 = tan-1

 = 37.761o

  38o

v = kmh-1

(c)The aeroplane has to do another drop but from a height 180m. What is the maximum speed (in kmh-1) that the plane can have for the food to land within 300m from the hiker if the parcel is again released when the plane is over the hiker?

In x direction

v = ux= ? d= 300mt = ?

v=

v=

v = 50ms-1

v = 180kmh-1

x

d = 300m

Find t from the y direction

t = ? u= 0 x= 180m a= 10ms-1

x= ut + ½ at2

180= ½ × 10 × t2

180 = 5 × t2

36 = t2

6 = t

y

6s

v = kmh-1

(c)The aeroplane has to do another drop but from a height 180m. What is the maximum speed (in kmh-1) that the plane can have for the food to land within 300m from the hiker if the parcel is again released when the plane is over the hiker?

In x direction

v = ux= ? d= 300mt = ?

v=

v=

v = 50ms-1

v = 180kmh-1

x

d = 300m

Alternative using derived formulae

u = ? Rhproj = 300 h = 180ms-1g = 10ms-2

Rhproj = u

300=

300= × 6

50 = u

u = 180kmh-1

Find t from the y direction

t = ? u= 0 x= 180m a= 10ms-1

x= ut + ½ at2

180= ½ × 10 × t2

180 = 5 × t2

36 = t2

6 = t

y

6s

A car is driven off roof of a 15m building at 10ms-1 and lands on an adjacent rooftop 5.0m away? How high is the adjacent building?

Find h2 from x direction

x =? u = 0 a = 10ms-2t = ?

x= ut + ½ at2

x = ½ × 10 × 0.52

x= 1.25m

Find height of second building

Height = 15 – 1.25

= 13.75

 14m

x

h1

h2

Not drawn to scale

15m

5.0m

20ms-1

0.5s

Find t from the xdirection

t = ? v= ux= 10ms-1d= 5.0m

t=

t=

t = 0.5s

y

A car is driven off roof of a 15m building at 10ms-1 and lands on an adjacent rooftop 5.0m away? How high is the adjacent building?

Find h2 from x direction

x =? u = 0 a = 10ms-2t = ?

x= ut + ½ at2

x = ½ × 10 × 0.52

x= 1.25m

Find height of second building

Height = 15 – 1.25

= 13.75

 14m

x

h1

h2

Not drawn to scale

15m

5.0m

Alternative using derived formulae

h = h2 = ? Rhproj = 5.0 u = 10ms-1g = 10ms-1

Rhproj = u

5 =

0.5=

10ms-1

0.5s

Find t from the xdirection

t = ? v= ux= 10ms-1d= 5.0m

t=

t=

t = 0.5s

y

0.25 =

1.25 = h

Find height of second building

Height = 15 – 1.25

= 13.75

 14m

Horizontal Projectile Motion

Exam Questions

A sportscar is driven horizontally off building 1 and lands on a building 20m away. The floor where the car lands in building 2 is 4.0 m below the floor from which it started in building 1.

Calculate the minimum speed at which the car should leave building 1 in order to land in the car park of building .

In x direction

v = ux= ? d= 20mt= ?

v=

v=

v = 22.3607ms-1

v  22 ms-1

0.89443s

Find t from the y direction

t = ? u= 0 x= 4m a= 10ms-2

x= ut + ½ at2

4= ½ × 10 × t2

4= 5 × t2

0.8 = t2

0.89443 = t

A sportscar is driven horizontally off building 1 and lands on a building 20m away. The floor where the car lands in building 2 is 4.0 m below the floor from which it started in building 1.

Calculate the minimum speed at which the car should leave building 1 in order to land in the car park of building .

Alternative using derived formulae

u = ? Rhproj = 20.0 h = 4.0ms-1g = 10ms-2

Rhproj = u

20=

20= × 0.89443

22.3607 = u

u 22 ms-1

In x direction

v = ux= ? d= 20mt= ?

v=

v=

v = 22.3607ms-1

v  22 ms-1

0.89443s

Find t from the y direction

t = ? u= 0 x= 4m a= 10ms-2

x= ut + ½ at2

4= ½ × 10 × t2

4= 5 × t2

0.8 = t2

0.89443 = t

2002 Exam Q6Q2

In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of25 ms–1

Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2.

Find final velocity

h2 = a2 + b2

v2= 252+ 8.9944272

v2= 705

v= 26.5518

v 27 ms-1

25ms-1

8.94427ms-1

v

In y direction

vy = ? uy= 0 x=4.0m a = 10ms-2

v2 = u2 + 2ax

v2 = 2 × 10 × 4

v2=80

v = 8.94427

2002 Exam Q6Q2

In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of25 ms–1

Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2.

Find final velocity

h2 = a2 + b2

v2= 252+ 8.9944272

v2= 705

v= 26.5518

v 27 ms-1

25ms-1

8.94427ms-1

38o

Alternative using Derived Formula

v = ? u = 25ms-1h = 4.0m g = 10ms-1

v =

v =

v = 26.5518

v  27ms-1

v

tan  =

tan  =

 = tan-1

 = 18.96173o

  19o

v

126 ms-1

In y direction

vy = ? uy= 0 x=4.0m a = 10ms-2

v2 = u2 + 2ax

v2 = 2 × 10 × 4

v2=80

v = 8.94427

tan  =

tan  =

 = tan-1

 = 37.761o

  38o

19o

27 ms-1

2002 Exam Q6Q2

In order to be sure of landing in the car park of building 2, Taylor and Jones in fact left building 1 at a speed of25 ms–1

Calculate the magnitude of the velocity of the car just prior to landing in the car park of building 2.

Find final velocity

h2 = a2 + b2

v2= 252+ 8.9944272

v2= 705

v= 26.5518

v 27 ms-1

25ms-1

8.94427ms-1

38o

Alternative using energy approach

v = ? u = 25ms-1h = 4.0m g = 10Nkg-1

Eki+ Ug Ekf

½mu2 + mgh = ½mv2

u2+ 2gh = v2

v =

v

tan  =

tan  =

 = tan-1

 = 18.96173o

  19o

v

126 ms-1

In y direction

vy = ? uy= 0 x=4.0m a = 10ms-2

v2 = u2 + 2ax

v2 = 2 × 10 × 4

v2=80

v = 8.94427

v =

v = 26.5518

v  27ms-1

tan  =

tan  =

 = tan-1

 = 37.761o

  38o

19o

Same as derived formula

27 ms-1

2006 Exam Q8Q3

A 0.5kg rocket is launched horizontally and propelled by a constant force of 22 N for 1.5 s, from the top of a 50 m tall building. Assume that in its subsequent motion the rocket always points horizontally.

After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to the ground?

x

In x direction

v= ? t= 1.5su = 0 a= = = 44ms-2

v = u + at

v= 0 + 44 × 1.5

v = 66 ms-1

In y direction

v = ? t = 1.5su = 0 a = 10ms-2

v = u + at

v = 0 + 10 × 1.5

v = 15 ms-1

22N

y

alternative using momentum/impulse approach

v = ? t = 1.5su = 0 Fnet = 22N

Ft= m(v – u)

22 × 1.5 = 0.5v

33 = 0.5v

66 ms-1 = v

2006 Exam Q8Q3

After 1.5 s, what is the speed of the rocket, and at what angle is the rocket moving relative to the ground?

x

Find final velocity

h2 = a2 + b2

v2= 662+ 152

v2= 4581

v= 67.6831

v 68 ms-1

66ms-1

15ms-1

22N

v

tan  =

tan  =

 = tan-1

 = 12.8043o

  13o

y

13o

68 ms-1

Obliquely Launched Projectile Motion

Obliquely Launched Projectile Motion

Case 1: landing height = launch height

Case 2: landing height different to launch height

Case 1: Obliquely Launched Projectiles Landing At The Launching Height

In x dirn

vx= u cosin

In y direction

uy = u sina = –10ms-1use constaccelequns

The motion in the x & y directions are independent

uy = u sin vxis constant = u cos

x

If not enough info in one dirn get time from the other dirn

y

Reference

axes

If time not involved

Eki Ekf+ Ug

½mu2= ½mv2+ mgh

u2= v2+ 2gh

vy

vx

vx

vx

vx

d

vy

vy

t

v

u

Ekmin = ½ mvx2

uy = u sin 

=

ux = ucos

Speeds before hmax are the same as those after at each height

v = u

Eight points

Case 1: Obliquely Launched Projectiles Landing At The Launching Height

hmax=

range=

tflight=

x

y

vy

vx

vx

vx

vx

vy

vy

u

uy = u sin 

ux = ucos

=

v = u

Three formulae

Case Launching Height2: Obliquely Launched Projectiles Landing At A Different Height To The Launching Height

x

Notes for Case 2

• The eight points relating to oblique projectiles that land at the launching height also apply to this type of situation.

• You will not be asked to find the time from the vertical axis since it will involve solving a quadratic equation. Instead you will either be given the time or asked to work it out from a horizontal distance (using t =

u

Obliquely Launched Projectile Motion Launching Height

Worked Examples

Obliquely Launched Launching HeightProjectile MotionExample 1

• What are the three other ways of working out the flight time?

• t= ? u = 13.101 x =0 a = –10ms-2 (this will involve a factorisation)

• t = ? u = 13.101 v = –13.101 a = –10ms-2 (this is pretty easy)

• Using flight time derived formula (just put the numbers in and work out)

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(a)How far will the ball travel before it strikes the ground if the ground is flat?

In x direction

d = ? v = 28.096ms-1t = ?

25o

uy =31sin25

= 13.101ms-1

x

y

ux =31cos25

= 28.096ms-1

d

Find t to top of path from y dirn

t = ? u = 13.101 v =0 a = –10ms-2

v= u + at

0 = 13.101 + – 10 × t

– 13.101 = – 10 × t

1.3101 = t

Find total time

Total time = 2 × 1.3101

= 2.6202s

Obliquely Launched Launching HeightProjectile MotionExample 1

• What are the three other ways of working out the flight time?

• t= ? u = 13.101 x =0 a = –10ms-2 (this will involve a factorisation)

• t = ? u = 13.101 v = –13.101 a = –10ms-2 (this is pretty easy)

• Using flight time derived formula (just put the numbers in and work out)

uy =31sin25

= 13.101ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(a)How far will the ball travel before it strikes the ground if the ground is flat?

In x direction

d = ? v = 28.096ms-1t = ?

ux =31cos25

= 28.096ms-1

x

y

Find t from y dirn

t = ? u = 13.101 x=-0 a = –10ms-2

x= ut+ ½at2

0 = 13.101 t + ½ × – 10 × t2

0 = 13.101 t– – 5 t2

0 = t (13.101 – 5t)

t = 0,

t= 2.6202

Obliquely Launched Launching HeightProjectile MotionExample 1

• What are the three other ways of working out the flight time?

• t= ? u = 13.101 x =0 a = –10ms-2 (this will involve a factorisation)

• t = ? u = 13.101 v = –13.101 a = –10ms-2 (this is pretty easy)

• Using flight time derived formula (just put the numbers in and work out)

uy =31sin25

= 13.101ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(a)How far will the ball travel before it strikes the ground if the ground is flat?

In x direction

d = ? v = 28.096ms-1t = ?

ux =31cos25

= 28.096ms-1

x

y

Find t from y dirn

t = ? u = 13.101 v =-13.101 a = –10ms-2

v= u + at

-13.101 = 13.101 + – 10 × t

– 26.202 = – 10 × t

2.6202 = t

Obliquely Launched Launching HeightProjectile MotionExample 1

• What are the three other ways of working out the flight time?

• t= ? u = 13.101 x =0 a = –10ms-2 (this will involve a factorisation)

• t = ? u = 13.101 v = –13.101 a = –10ms-2 (this is pretty easy)

• Using flight time derived formula (just put the numbers in and work out)

uy =31sin25

= 13.101ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(a)How far will the ball travel before it strikes the ground if the ground is flat?

In x direction

d = ? v = 28.096ms-1t = ?

d = vt

d = 28.096 × 2.6202

d = 73.617

d  74m

ux =31cos25

= 28.096ms-1

x

y

Find t from flight formula

t = ? u = 31ms-1= 25og = –10ms-2

tflight=

tflight=

tflight= 2.6202

2.6202s

Obliquely Launched Projectile Motion Launching HeightExample 1

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(a)How far will the ball travel before it strikes the ground if the ground is flat?

In x direction

d = ? v = 28.096ms-1t = ?

d = vt

d = 28.096 × 2.6202

d = 73.617

d  74m

25o

range

1 step alternative using derived formulae

Robliq = ? u = 31ms-1 = 25o g = 10ms-1

Robliq=

Robliq=

Robliq= 73.617

Robliq74m

Find t to top of path from y dirn

t = ? u = 13.101 v =0 a = –10ms-2

v= u + at

0 = 13.101 + – 10 ×

– 13.101 = – 10 × t

1.3101 = t

Find total time

Total time = 2 × 1.3101

= 2.6202s

2.6202s

Obliquely Launched Projectile Motion Launching HeightExample 1

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(b)What is the minimum speed of the golf ball?

Minimum speed occurs at the top of the path when there only the horizontal component of velocity.

So minimum is 28.096ms-1

25o

uy =31sin25

= 13.101ms-1

x

y

ux =31cos25

= 28.096ms-1

Obliquely Launched Projectile Motion Launching HeightExample 1

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(c)How high will the golf ball rise?

25o

uy =31sin25

= 13.101ms-1

x

y

ux =31cos25

= 28.096ms-1

In y direction

x= ? u = 13.101ms-1 v = 0 a= –10ms-2

v2= u2 + 2ax

0 = 13.1012+ 2 × – 10 × x

0 = 171.636 + – 20 × x

– 171.636= – 20 × x

8.5818 = x

x = 8.6m

Obliquely Launched Projectile Motion Launching HeightExample 1

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(c)How high will the golf ball rise?

25o

uy =31sin25

= 13.101ms-1

x

y

ux =31cos25

= 28.096ms-1

Alternative using derived formulae

hmax= ? u = 31ms-1 = 25o g = 10ms-1

hmax=

hmax=

hmax=8.5820

hmax8.6m

In y direction

x= ? u = 13.101ms-1 v = 0 a= –10ms-2

v2= u2 + 2ax

0 = 13.1012+ 2 × – 10 × x

0 = 171.636 + – 20 × x

– 171.636= – 20 × x

8.5818 = x

x = 8.6m

Obliquely Launched Projectile Motion Launching HeightExample 1

31ms-1

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(d)What is the velocity of the ball as it hits the ground?

Since the ball hits the ground at the same height as it launch height final speed is:

25o

25o

uy =31sin25

= 13.101ms-1

x

y

ux =31cos25

= 28.096ms-1

Obliquely Launched Projectile Motion Launching HeightExample 1

31ms-1

A golf ball is hit from the ground at 31ms-1 at 25o to the horizontal.

(e)What will be the speed of the ball 20m off the ground?

25o

x

y

Using energy approach (which avoids having to work with vectors)

v = ? u = 31ms-1h = 20m g = 10ms-1

Eki Ekf+ Ug

½mu2= ½mv2 + mgh

½u2= ½v2+ gh

u2= v2+ 2gh

312= v2+ 2×10×20

961 = v2 + 400

561 = v2

23.685 = v

v  24ms-1

Obliquely Launched Projectile Motion Launching HeightExample 2

25ms-1

x

A golf ball is hit at 25ms-1 at 35o to the horizontal from a cliff edge and it takes 6.0 seconds to hit the ground below the cliff.

(a) How high is the cliff?

In y direction

x= ? u= 14.3390ms-1t = 6.0s a = –10ms-2

x = ut + ½ at2

x= 14.3390× 6 + ½ × – 10 × 62

x= – 93.966

so the cliff is  94m high

35o

x

y

uy =25sin35

= 14.339ms-1

ux =25cos35

= 20.4796ms-1

Obliquely Launched Projectile Motion Launching HeightExample 2

25ms-1

A golf ball is hit at 25ms-1 at 35o to the horizontal from a cliff edge and it takes 6.0 seconds to hit the ground below the cliff.

(b) How far from the base of the cliff does the ball land?

In x direction

d = ? v = 20.4796ms-1t = 6

d = vt

d = 20.4796× 6

d = 122.875

d  123m

35o

x

y

d

uy =25sin35

= 14.339ms-1

ux =25cos35

= 20.4796ms-1

Obliquely Launched Projectile Motion Launching HeightExample 3

28ms-1

28ms-1

h

A golf ball is hit at 28ms-1at 60oto the horizontal from a fairway and hits a tree that is 56m away. How far up the tree does the golf ball strike the tree?

In y direction

x= ? u = 24.249ms-1a= –10ms-2t = ?

x = ut + ½ at2

x = 24.249× 4+ ½ × – 10 × 42

x = 16.996

so the ball hits the tree  17m up

60o

x

60o

y

56m

uy =28sin60

= 24.249ms-1

ux =28cos60

= 14ms-1

Find t from the xdirection

t = ? v= ux= 14ms-1d= 56m

t=

t=

t = 4 s

4s

Obliquely Launched Projectile Motion Launching Height

Exam Questions

2004 Launching HeightExam Q7Q1

The diagram shows a motorcycle rider using a 20° ramp to jump her motorcycle across a river that is 10.0 m wide.

Question 7

Calculate the minimum speed that the motorcycle and rider must leave the top of the first ramp to cross safely to the second ramp that is at the same height. (The motorcycle and rider can be treated as a point-particle.)

u = ? Robliq =10  = 20og = 10ms-2

Robliq=

10 =

100 = u2sin40

155.572 = u2

12.4729 = u

u 12 ms-1

Using x & y direction analysis for this problem is very difficult so using the range formula is a much more practical way to solve this problem

2005 Launching HeightExam Q11Q2

8o

30ms-1

3.0m

A ball leaves a racket 3.0 m above the ground at atan angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s-1.

With what speed, relative to the deck, did the ball leave Fred's racket? Give your answer to three significantfigures.

x

y

30ms-1

cos  =

cos8=

u =

u = 30.29483ms-1

u 30.3ms-1

u

8o

30ms-1

2005 Launching HeightExam Q12Q3

x

8o

30ms-1

3.0m

3.0m

A ball leaves a racket 3.0 m above the ground at atan angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s-1.

At its highest point, how far was the ball above the ground.

x= ? u= 30ms-1v = 0 a= –10ms-2

v2= u2 + 2ax

0 = 4.2162252+ 2 × – 10 × x

0 = 17.77655 + – 20 × x

– 17.77655 = – 20 × x

0.888828 = x

Overall height = 0.888828 + 3

= 3.888828

 3.89 m

x

y

30ms-1

30.29483ms-1

30.29483sin8

= 4.216225ms-1

8o

30ms-1

2005 Launching HeightExam Q12Q3

hmax

8o

30ms-1

3.0m

3.0m

A ball leaves a racket 3.0 m above the ground at atan angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s-1.

At its highest point, how far was the ball above the ground.

x= ? u= 30ms-1v = 0 a= –10ms-2

v2= u2 + 2ax

0 = 4.2162252+ 2 × – 10 × x

0 = 17.77655 + – 20 × x

– 17.77655 = – 20 × x

0.888828 = x

Overall height = 0.888828 + 3

= 3.888828

 3.89 m

x

y

30ms-1

Alternative using derived formulae

hmax= ? u = 30.29483ms-1 = 8o g = 10ms-2

hmax=

hmax=

hmax= 0.888828

30.29483ms-1

8o

30ms-1

2005 Launching HeightExam Q12Q3

8o

h2

h1 =

30ms-1

3.0m

A ball leaves a racket 3.0 m above the ground at atan angle of 8° to the horizontal. At its maximum height it has a speed of 30.0 m s-1.

At its highest point, how far was the ball above the ground.

x= ? u= 30ms-1v = 0 a= –10ms-2

v2= u2 + 2ax

0 = 4.2162252+ 2 × – 10 × x

0 = 17.77655 + – 20 × x

– 17.77655 = – 20 × x

0.888828 = x

Overall height = 0.888828 + 3

= 3.888828

 3.89 m

x

y

30ms-1

Alternative using Energy Approach

h2= ? u = 30.29483ms-1h1 = 3m g = 10ms-2

Eki + Ugi Ekf+ Ugf

½ mu2 + mgh1 = ½ mv2+ mgh2

½ u2+ gh1= ½ v2+ gh2

½ × 30.294832 + 10 × 3 = ½ × 302+ 10 × h2

488.888 = 450 + 10 × h2

3.8888 = h2

h2 3.89m

30.29483ms-1

8o

30ms-1

2007 Launching HeightExam Q14Q4

40ms-1

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

What is the time of ﬂight of the paintball?

25o

x

y

Find t from the xdirection

t = ? v= ux= 28.096ms-1d= 127m

t=

t=

t = 3.5032 s

t 3.5 s

uy =40sin25

= 16.9047ms-1

ux =40cos25

= 36.2523ms-1

2007 Launching HeightExam Q14Q4

40ms-1

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

What is the time of ﬂight of the paintball?

25o

x

y

Alternative

Find t from the ydirection

t = ? u= 16.9047ms-1v= –16.9047ms-1a = –10ms-2

v= u + at

-16.9047 = 16.9047 + – 10 × t

– 33.8094 = – 10 × t

3.38094 = t

t  3.4 s

uy =40sin25

= 16.9047ms-1

ux =40cos25

= 36.2523ms-1

There is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction

2007 Launching HeightExam Q14Q4

40ms-1

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

What is the time of ﬂight of the paintball?

25o

x

y

Find t from the ydirection

t = ? u= 16.9047ms-1v= –16.9047ms-1a = –10ms-2

v= u + at

-16.9047 = 16.9047 + – 10 × t

– 33.8094 = – 10 × t

3.38094 = t

t  3.4 s

uy =40sin25

= 16.9047ms-1

ux =40cos25

= 36.2523ms-1

There is a discrepancy in the data for this problem hence the slight difference in answers from the x direction and y direction

2007 Launching HeightExam Q14Q4

40ms-1

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

What is the time of ﬂight of the paintball?

25o

x

y

Alternative with formula

Find t from flight formula

t= ? u = 40ms-1= 25og = –10ms-2

tflight=

tflight=

tflight= 3.3809

tflight3.4s

uy =40sin25

= 16.9047ms-1

ux =40cos25

= 36.2523ms-1

There is a discrepancy in the data for this problem hence the slight difference in answers from the previous methods

2007 Launching HeightExam Q15Q5

40ms-1

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

What is the value of h, the maximum height above the ﬁring level?

25o

x

y

Find hmaxfrom y direction

x = ? u = 16.9047ms-1 v = 0 a = –10ms-2

v2 = u2 + 2ax

0 = 16.90472+ 2 × – 10 × x

0 = 287.201+ – 20 × x

– 287.201= – 20 × x

14.3600 = x

x 14m

uy =40sin25

= 16.9047ms-1

ux =40cos25

= 36.2523ms-1

2007 Launching HeightExam Q15Q5

40ms-1

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

What is the value of h, the maximum height above the ﬁring level?

25o

x

y

Alternative with formula

Find t from flight formula

hmax= ? u = 40ms-1 = 25o g = 10ms-2

hmax=

hmax=

hmax=14.2885

hmax14m

uy =40sin25

= 16.9047ms-1

ux =40cos25

= 36.2523ms-1

There is a discrepancy in the data for this problem hence the slight difference in answers from the previous method

2007 Launching HeightExam Q16Q6

Daniel ﬁres a ‘paintball’ at an angle of 25° to the horizontal and a speed of 40.0 ms–1. The paintball hits John, who is 127 m away. The height at which the ball hits John and the height from which the ball was ﬁred are the same.

Which of the following diagrams (A–D) below gives the direction of the force acting on the paintball at pointsX and Y respectively?

Since air resistance is ignored the only force on the paintball is gravity which is constant and downwards, so the acceleration must be constant and downwards.

2007 Launching HeightExam Q17Q7

Later in the game, Daniel is twice as far away from John (254 m). John ﬁres an identical paintball from the same height above the ground as before. The ball hits Daniel at the same height as before. In both cases the paintball reaches the same maximum height (h) above the ground.

Which one or more of the following is the same in both cases?

A. ﬂight time

B. initial speed

C. acceleration

D. angle of ﬁring level?

Because the paintball gets to the same vertical height the vertical component in each situation must be the same and hence the flight time will also be the same.

Since air resistance is ignored the only force on the paintball is gravity which is constant in both situations and so the acceleration is constant as well.

So With More Than One Method For Solving Most Projectile Motion Questions – How Do I Decide On The Best?

Some Loose Rules To Follow.

• The best method is the one that makes the most sense to you and doesn’t take forever to get to an answer

• Often if there is a formula use a formula, but be careful of expressions such as

• hmax==0.888828

• Robliq=

• v = = 126.491

• Rhproj = = 774.597

• Make sure you practice more than one method when solving projectile motion questions – particularly working with x & y directions.

• Constant Accel Equations Problem

Some Loose Rules For Deciding On The Best Method To Solve A Problem

Energy Formulae

(no time)

Impulse/ Momentum Formulae

(no dist)

• An energy approach is very useful when asked for the magnitude of a velocity that is not at launching height or maximum height.

• Remember there are three approaches that can be used for solving