Question

A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration rate is -3.7m/s². Find the velocity with which the camera hits the ground.

Answer 1

42.1m/s

Step-by-step explanation:

Given parameters:

Height of cliff = 239m

Free fall acceleration = -3.7m/s²

Unknown:

Final velocity of the camera = ?

Solution:

When the body is falling under free fall, the acceleration = 3.7m/s² ;

The appropriate motion equation is:

V² = U² + 2gH

V = final velocity

U = initial velocity

g = free fall acceleration

H = height of the fall

Insert the parameters and solve;

U = 0

V² = 0² + 2 x 3.7 x 239

V² = 1768.6

V = 42.1m/s