Projectiles

Practice

practice problem 1

In the 1994 action-adventure film Speed, an extortionist equipped a Los Angeles bus with a bomb that was set explode if the speed of the bus fell below 50 mph (22 m/s). The police discovered the bomb and routed the bus on to a segment of freeway that was still under construction — their intention being to keep it out of the notoriously heavy Southern California traffic. In a twist of the plot, however, they overlooked a 50 foot (15 m) gap in the construction. Since the bomb would explode killing everyone on board if they slowed down, they decide to jump the gap. The missing segment lies on an effectively level stretch of elevated freeway, 50 feet (15 m) above the ground, yet somehow they manage to jump the gap at 67 mph (30 m/s) and land safely on the other side.
     
     
  1. If the bus in Speed obeyed the laws of physics how would this scene end?
  2. Just out of curiosity, what would happen if the bus in the movie Speed was driven off of a 15 m high, horizontal ramp?
    1. Where would the bus land?
    2. What velocity would it have when it struck the ground?

solution

  1. This question is basically asking how far down a bus moving at 30 m/s would fall in the time it took it to move forward 15 m. Although the bus is accelerating downward due to gravity its horizontal velocity remains constant. (Gravity never acts horizontally.) It will take the bus …
             
    Δt =  Δx  =  15 m  = 0.50 s
    v 30 m/s

    to travel the width of the gap, during which time it would fall …

    y =  1  at2 =  1  ( 9.8 m/s2)(0.50 s)2 = 1.3 m
    2 2
             
    This means the bus will strike the flat concrete face of the freeway stub somewhere above its front bumper. So much for Speed II.
  2. Let's let the bus fly free!
    1. The first half of this question is basically asking how far forward a bus moving at 30 m/s would travel in the time it took for it to fall 15 m downward. In this problem there are two independent equations of motion -- one with constant velocity (the horizontal motion) and one with constant acceleration (the vertical motion). Since the ramp is horizontal, there is no initial velocity in the vertical direction. Thus, the time it takes a horizontally launched projectile to reach the ground is the same as the time it takes an object released from rest to fall the same height.
               
      y0 =  0 m   y =  y0 + v0yΔt + ½ayΔt2
      y =  15 m   y =  ½ayΔt2
      v0y =  0 m/s   Δt =  √(2y/ay) = √[2(15 m)(9.8 m/s2)]
      ay =  9.8 m/s2   Δt =  1.75 s

      Now use the time of flight to determine the horizontal displacement of the bus as it flies through the air.

      vx =  30 m/s   x =  vxΔt
      Δt =  1.75 s   x =  (30 m/s)(1.75 s)
            x =  53 m
               
      Look for the bus 53 m in front of a point directly below the edge of the ramp.
    2. The final velocity of the bus will be the vector sum of its horizontal and vertical velocities on impact. Since there is no horizontal acceleration, the velocity in this direction remains constant. If the bus leaves the ramp at 30 m/s horizontally it will strike the ground at 30 m/s horizontally. There is an acceleration vertically, however. Initially, the bus has no vertical velocity, but this will obviously change. Gravity pulls everything relentlessly toward the earth. Once there's no more road to hold up the bus, there's nothing to keep it from accelerating down.
               
      y0 =  0 m   vy2 =  voy2 + 2ayy
      y =  15 m   vy =  √(2ayy)
      v0y =  0 m/s   vy =  √[2(9.8 m/s2)(15 m)]
      ay =  9.8 m/s2   vy =  17.1 m/s

      Now that we have both components, find the magnitude and direction of the resultant velocity by the usual means: Pythagorean theorem …

      v2 =  vx2 + vy2
      v2 =  (30 m/s)2 + (17.1 m/s)2
      v =  35 m/s

      and tangent …

      tan θ =  vy  =  17.1 m/s  
      vx 30 m/s  
        θ =  30° (angle of depression)
       

practice problem 2

Projectile launcher demo

solution

Answer it.

practice problem 3

Shoot the monkey demo

solution


[magnify]

Describe the solution.

practice problem 4

  1. max range 45°
  2. max height 90°
  3. max flight time 90°
  4. max path length 56.46°
  5. max area under curve 60°

solution

Answer it.

The Physics Hypertextbook

  • No condition is permanent.