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Projectile motion. ballistic motion free fall. Assumptions. The Earth’s surface is flat (a plane) Gravitational acceleration g = 9.81 m/s 2 There is no air resistance ( Later on during the course, we will return to the subject with more general assumptions ).

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Projectile motion

Projectile motion

ballistic motion

free fall


Assumptions
Assumptions

  • The Earth’s surface is flat (a plane)

  • Gravitational acceleration g = 9.81 m/s2

  • There is no air resistance

    ( Later on during the course, we will return to the subject with more general assumptions )


Approximate solution without a calculator:

Answers: Flight time is 8 sec.

Highest point 80 m


The bullet experiences 2 different types of motions . Horizontally it is in a uniform motion with a constant speed. Vertically its motion is uniformly accelerating with a = - 9.81 m/s2.

1) Divide the initial velocity vector into horizontal and vertical components:

v0 sinα=229.4

v0=400 m/s

v0 cosα=400*cos35o =327.7 m/s


2. Calculate the flight time. m/s in 35 deg angle.

Vertical velocity is initially 229.4 m/s. It changes with -9.81 every second. => After about 23 seconds velocity is zero and the bullet is at the ceiling point. After about 46 seconds the bullet hits the ground.

Mathematically: The equation for vertical velocity is:

vy = v0 sinα – g t

Substituting vy = 0 we get the time when the bullet is at highest.

t = v0 sinα / g ( here 23.4 s)

3. Calculate the highest point

Vertical velocity is initially 229.4 m/s. At the ceiling point it is 0. The average velocity during the rise is 229.4/2. The highest y- coordinate is 229.4/2 m/s* 23.4 s = 2684 m = 2. 6 km.

Mathematically: The highest point

ymax = v0 sinα / 2* v0 sinα / g

=> ymax = v02 (sinα)2 / (2g)


3. Calculate the horizontal distance of the flight m/s in 35 deg angle.

Horizontal velocity is initially and all the time 327.7 m/s.

Once we know the flight time 46.8 s , it is easy to calculate the distance x = 327.7 m/s * 46.8 s = 15336 m = 15.3 km

  • Mathematically: The horizontal velocity is v0 cosα

  • and the flight time = 2 v0 sinα / g

    • x = 2 v02 sin α cos α / g

      x = v02/g sin(2α)

Observations: The horizontal distance is proportional to the square of the initial velocity.

sin(2α) = 1 when α = 45o.

The shot or throw is longest, when the angle is 45 degrees.

( This is true only when the starting and landing levels are same)


The equations for x y
The equations for (x,y) m/s in 35 deg angle.

In Olympic games a sportsman throws hammer, which starts its flight at 1.8 m height at the speed of 27.0 m/s. The angle is 47 degrees. Calculate the result:

The flight time:

Result:


With mathematica
With Mathematica m/s in 35 deg angle.


Problem: Olympic winner of shot put threw 21.32. Assume that the angle was 46 degrees and the ball started at 1.8 m height.

Calculate the initial velocity of the throw.

Result: The initial velocity v0 was 13.91 m/s

and the time = 2.206 s


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