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# Tutorial for Projectile Motion Problems - PowerPoint PPT Presentation

Tutorial for Projectile Motion Problems. Components of Velocity. If an object is launched at an angle, its velocity is not perfectly horizontal (x direction), nor is it perfectly vertical (y direction).

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## PowerPoint Slideshow about ' Tutorial for Projectile Motion Problems' - jolanta-dbrowski

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### Tutorial for Projectile Motion Problems

• If an object is launched at an angle, its velocity is not perfectly horizontal (x direction), nor is it perfectly vertical (y direction).

• An object launched at an angle will have a velocity that is a combination of the Vx and the Vy.

• If the velocity is given, you must use the angle to determine what part of the initial velocity is in the X direction and what part is in the Y direction.

• V0y=V0sin Θ (this is the equation for V0y always)

• Vx=V0 cos Θ

• These are known as the “components” of the velocity.

• The velocity in the Y direction must change. The velocity in the Y direction decreases because any object that goes up must be affected by gravity.

• The velocity in the X direction does not change at all, it is the same EVERYWHERE along the path.

• The time it takes for an object to reach its maximum height will be half of the time it spends in the air.

• At the maximum height, the velocity in the Y direction will be 0 m/s

• The total time an object is in the air is simply twice the half time.

• The total time is used to find the maximum range (X) of the object (measured in meters).

• To find maximum height, use the following equation:

• The time in this equation must be the half time!

• To find the maximum range, you must use the total time in the following equation:

• If the equation calls for V0y, you must use the V0y value you calculated using V0sinΘ.

• If the equation calls for Vx, you must use the Vx value you calculated using V0cos Θ.

• Please note that Vx, V0y, and V0 are not the same value!!!

• If you are asked to find the V1 at a specific time, you must first find the components of the initial velocity (Vx and V0y).

• Next, find the V1y at the given time using the below equation:

• Once you have V1y at the given time, and you have Vx, simply use the Pythagorean Theorem to find V1:

• To find the angle at any point, first find the Vx and the Vy at that point.

• Once those are determined, use the tangent trig function.

• Θ=Tan-1 (Vy/Vx)

• A ball is kicked into the air at 30 m/s at an angle of 25 degrees. What is its maximum range and height?

Vx=V0 cos Θ=30 cos 25= 27.19 m/s

V0y=V0 sin Θ= 30 sin 25=12.68 m/s

Given:

V1: 30 m/s

Θ=25 degrees

V2y=0 m/s

g= -9.8 m/s2

Find:

Vx

V1y

th

tt

X

Y

Ttotal=2.59 s

A rocket is launched from the ground and is aimed 2500 m away. If it reaches a maximum height of 200 meters, how fast must it be going when it leaves the ground (find Vy and Vx)? At what angle must it be launched? (Hint: find time first using cliff problems)

Given:

X: 2500 m

Y: 200 m

Find:

V1y

Vx

th

tt

V1

To find the initial velocity (V0), find Vx and V0y. First find time using the initial velocity in the y direction as zero.

= 6.39s

T Total=12.78s

Vx= x = 195.62 m/s

t

V0y= V1y-gt

= 62.62 m/s

3. A basketball player jumps up at 15 m/s at an angle of 50 degrees. At what time will he be 2 meters away (horizontally) from where he jumped?

Vx=V0 cos Θ=9.64 m/s

T=x= .21 s

Vx

Given:

V0: 15 m/s

Θ=50 degrees

X=2 m/s

Find:

t

4. A football is kicked with an initial velocity of 55 m/s at an angle of 60 degrees. At what times will it pass 5 meters above the ground? (you can use quadratic formula in the calculator)

V0y= V0 sin Θ= 47.63 m/s

Given:

V0: 55 m/s

Θ=60 degrees

Y=5 m

Find:

T1

T2

The ball will pass 2 meters high twice.

You can use the quadratic formula to find the two times

ax2+ bx + c=0

T1=.11s

T2= 9.61s

Make equation look like this

1/2gt2+V1yt-y=0

Θ

5. A baseball is thrown into the air with a velocity of 27.5 m/s at an angle of 45 degrees. What will its velocity be after 3 seconds? (Hint: find its Vy and Vx at 3 seconds, use the Pythagorean Theorem).

Given:

V0: 27.5 m/s

Θ=45 degrees

T= 15 sec.

Find:

V1

V1y

Vx

V0y

Vy= V0 sin Θ=19.45 m/s

Vx= V0 cos Θ= 19.45 m/s

V1y=V0y+gt=9.65 m/s

6. A rocket is shot up into the air. If it has a range of 1000 m, and reaches a height of 800 m, what angle was it shot up at and what was its initial velocity?

You must first find the half way time, how long it the rocket to drop the 800 m; set V1y equal to zero for this.

Find:

V0

V0y

Vx

th

tt

Givens:

X=1000 m

Y= 800 m

G=9.8 m/s2

Next, find the total time by doubling the half way time

Ttotal= 25.56 seconds

Find Vx using the given distance (x) and the total time

#6 Continued 1000 m, and reaches a height of 800 m, what angle was it shot up at and what was its initial velocity?

Find V1y by setting V2y equal to zero and using the half way time

Find the V1 by using the Pythagorean Theorem with V1y and Vx.

Find the angle that it is launched at by using the tangent function

7. At what times will a rocket launched at an angle of 37 degrees and a velocity of 400 m/s pass 100 m high? Use Quadratic formula for this.

Find V1y first:

Given:

Θ=37 degrees

V0= 400 m/s

Y=100 m

Find:

t1

t2

V1y

To find the two times that the rocket passes that height, you must use the Quadratic formula: Set y=V0yt+1/2 gt2 to look like ax2+ bx+c=0. You are solving for t, so your A value is -4.9, your B value is 240.73 m/s (V0y), and your C value is -100 . Use the calculator to plug in and find the two times it passes those two heights.

T1= .42 s

t2= 48.71 s