During a walk on the Moon, an astronaut accidentally drops his camera over a 20.0-m cliff. It leaves his hands with zero speed, and after 2.00 s it attains a velocity of 3.40 m/s downward. How far has the camera fallen after 9.10 s

Δy=70.3885 downwards

Explanation:

First we get the acceleration on moon by substituting the values we have(v=0m/s at t=0s and v=-3.40m/s at t=2.0 s) into equation of simple motion:

As we know that

Δy=vit+(1/2)at²

Substitute the values of acceleration and time t=9.10s

So

Δy=vit+(1/2)at²

Δy=70.3885 downwards