In a stunt being filmed for a movie, a sports car overtakes a truck towing a ram

Question

In a stunt being filmed for a movie, a sports car overtakes a truck towing a ramp, drives up and off the ramp, soars into the air, and then lands on top of a flat trailer being towed by a second truck. The tops of the ramp and the flat trailer are the same height above the road, and the ramp is inclined 13° above the horizontal. Both trucks are driving at a constant speed of 15 m/s, and the flat trailer is 12 m from the end of the ramp. Neglect air resistance, and assume that the ramp changes the direction, but not the magnitude, of the car's initial velocity. What is the minimum speed the car must have, relative to the road, as it starts up the ramp?

Explanation / Answer

Lets assume that the trucks are stationary and road is moving between them with a velocity of u = -15 m/s.

Now, we can apply a simple range equation for car between the trucks which is given by

x = V²sin(2) / 2g

Here x = 12m, V = velocity of car relative to the trucks and is unknown

= 13o, g = 9.8 m/s2

Using these values in above equation we get

V = 23.16 m/s

Velocity of car relative to the road is v = V - u = 23.16 - (-15)

v = 38.16 m/s

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