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 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 11 m/s, and the trailer is 20 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

The horizontal distance traveled by the car = Range, R = 20 m The angle above the horizontal, = 13 degrees Let suppose the trucks are not moving We have a formula for the horizontal range, R = u^2sin2/g         u = sqrt(Rg/sin2) = sqrt(20*9.8/sin36) = 18.3 m/s Now we know that the trucks are moving with velocity 11 m/s So the total speed = 18.3 + 11 = 29.3 m/s So the minimum speed the car must have, relative to the road, as it starts up the ramp = 29.3 m/s

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