practice problem 1
- If the bus in Speed obeyed the laws of physics how would this scene end?
- Just out of curiosity, what would happen if the bus in the movie Speed was driven off of a 15 m high, horizontal ramp with nothing but the ground to land on?
- Where would the bus land?
- What velocity would the bus have when it struck the ground?
- Let's modify this stunt so that a bus like the one in the movie could jump over a 50 foot (15 m) gap in the freeway.
- How long does it take to jump the gap traveling at the speed shown in the movie?
- What upward velocity would a bus need to stay in the air this long?
- What direction should the bus head?
- What type of device is needed to make the bus travel in this direction?
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 v Δt = 15 m 30 m/s Δt = 0.50 s
to travel the width of the gap, during which time it would fall…
y = 1 at2 2 y = 1 (9.8 m/s2)(0.50 s)2 2 y = 1.3 m
This means the bus will strike the flat concrete face of the freeway stub somewhere above its front bumper. So much for Speed II.
- Let the bus fly free!
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 = 15 m v0y = 0 m/s ay = 9.8 m/s2 ∆t = ? y = y0 + v0yΔt + ½ayΔt2 y = ½ayΔt2 Δt = √(2y/ay) Δt = √[2(15 m)(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 Δt = 1.75 s x = ? x = vxΔt 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.
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 y = 15 m v0y = 0 m/s ay = 9.8 m/s2 vy = ? vy2 = voy2 + 2ayy vy = √(2ayy) vy = √[2(9.8 m/s2)(15 m)] 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
tan θ = vy = 17.1 m/s vx 30 m/s θ = 30° (angle of depression)
- ANSWER THIS: Let's modify this stunt so that a bus like the one in the movie could jump over a 50 foot (15 m) gap in the freeway.
- ANSWER THIS: How long does it take to jump the gap traveling at the speed shown in the movie?
- ANSWER THIS: What upward velocity would a bus need to stay in the air this long?
- ANSWER THIS: What direction should the bus head?
- ANSWER THIS: What type of device is needed to make the bus travel in this direction?
practice problem 2
- launch horizontally from a convenient height above the floor (to a reproducible location)
- measure the horizontal distance traveled
- measure the vertical height fallen
- calculate the time to fall
- calculate the horizontal velocity
- launch from an interesting angle above the horizontal
- measure the angle of inclination
- determine the components of the projectile's initial velocity
- determine the time to reach the highest point
- determine the height of the highest point (above the launch height)
- determine the location where the projectile returns to its launch height
practice problem 3
Describe the solution.
practice problem 4
- max range 45°
- max height 90°
- max flight time 90°
- max path length 56.46°
- max area under curve 60°