Question

On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the acceleration due to gravity on this planet?  m/s2 What is the final velocity of the rock just before it hits the planet's surface?  m/s

Answer 1

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

48.4 s = sqrt (2 (1.10e+02 m)/ g)

G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)

V = 4.523 m/s