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 17° above the horizontal. both trucks are driving at a constant speed of 17 m/s, and the flat trailer is 19 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?

The minimum speed of the car relative to the road must be 35 m/s.

Explanation:

The position of the flying car is given by the vector position r:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal position

v0 = initial velocity

t = time

α = angle of inclination of the ramp

y0 = initial vertical position

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the edge of the ramp.

From the figure, we can see that at final time the vector r is:

r final = (19 m, 0)

Then, using the equations for the x and y-components of the vector r, we can calculate the time of flight and the initial velocity of the car:

x = x0 + v0 · t · cos α

19 m = 0 m + v0 · t · cos α

solving for v0:

19 m / t · cos α = v0

Replacing (v0 = 19 m / t · cos α) in the equation of the y-component:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 m = 19 m/ (t ·cos α)  · t · sin α + 1/2 · g · t²

0 m = 19 m · tan α + 1/2 · g · t²

0 m = 19 m · tan 17° - 1/2 · 9.8 m/s² · t²

-19 m · tan 17°/ -4.9 m/s² =  t²

t = 1.1 s

Then, v0 will be:

v0 = 19 m / t · cos α

v0 = 19 m / 1.1 s · cos 17°

v0 = 18 m/s

This velocity is relative to the ramp. Since the ramp is moving at 17 m/s relative to the road, then, the velocity of the car relative to the road will be 17 m/s + 18 m/s = 35 m/s

12000j is work done