Question

A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .

Answer 1

a. We can find the velocity when the camera hits the ground. v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera hits the ground with a velocity of 42 m/s b. We can find the time it takes for the camera to hit the ground. y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds
       
It takes 11.4 seconds for the camera to hit the ground.