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.

42.1m/s

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

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

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