Question

A movie stunt driver on a motorcycle speeds horizontally off a 50.0 m high cliff. How fast must the motorcycle leave the cliff top to land on a level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.

Answer 1

Easiest way to go about this is to find the time it takes to get to the ground, then find how fast he must go to travel 90.0 m in that time:
To find the time taken, we use the equation of motion at constant acceleration in the y-direction only:
\[y=y_0+v_0t+at^2\]
Where y is the final height (which is 0 m), y0 is the starting heigh (which is 50.0 m), v0 is the starting velocity (which is 0 m/s), and a is the acceleration (which is -g), so now we have:
\[0\ m=(50.0\ m)-gt^2\]
Rearranging for t, we get:
\[t=\sqrt{\frac{50\ m}{g}}\approx 2.26\ s\]
Now, in the x-direction, due to there being no air-resistance, there is no acceleration, so we have the equation of motion at constant velocity in the x-direction:
\[x=x_0+vt\]
Where x is the final position (which is 90.0 m), x0 is the initial position, t is the time taken between the motorbike coming off the cliff and the motorbike hitting the ground, and v is the velocity at which he needs to go. Substituting, we get:
\[90.0\ m=0\ m+v(2.26\ s)\]
Rearranging for v, we get:
\[v=\frac{90\ m}{2.26\ s}\approx 39.8\ m\ s^{-1}\]
Checking our answer, this is about 150 km/h, which is quite reasonable for a motorbike to do. Also, we were given 3 significant figures for bothe the height of the cliff and the distance required to travel, and our answer is in 3 significant figures. So 39.8 m/s is your final answer

Answer 2

thank you so much :)