How to Calculate the Speed of a Stunt Car Leaving a Cliff

Question:

A stunt car goes speeding horizontally off a cliff that is 85 meters high. If the car lands 25 meters from the base of the cliff, how fast was the car driving as it left the cliff?

A) 3.4 m/s

B) 6.0 m/s

C) 37 m/s

D) 25 m/s

Answer:

To find the car's initial speed as it left the cliff, we can use the principles of projectile motion. By analyzing the horizontal and vertical motions separately, we can solve for the initial velocity. For this particular scenario, the car was driving at approximately 37 m/s.

Explanation:

To find the speed at which the car was driving as it left the cliff, we can use the principles of projectile motion. We can assume that the car's motion is solely in the horizontal direction, meaning there is no vertical acceleration. Using the equation d = v0t + 0.5at^2, where d is the horizontal distance traveled, v0 is the initial horizontal velocity, t is the time of flight, and a is the horizontal acceleration, we can solve for v0. The initial vertical velocity v0y is 0, since the car is not initially moving in the vertical direction. We can use the equation h = v0yt + 0.5gt^2, where h is the height of the cliff, g is the acceleration due to gravity, and t is the time of flight, to solve for t. Plugging in the given values, we find that the car's initial velocity v0 is approximately 37 m/s.