Question

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet.
Compared to the acceleration due to gravity near Earth’s surface, the acceleration due to gravity near the
surface of the planet is approximately

Answer 1

On the planet that is NOT earth,
Y (1/2)g'*t^2
22 = (g'/2)*(3.0)^2
so the value of the acceleration of gravity there is
g' = 44/9 = 4.89 m/s^2

The is just about half the value of g at the surface of the Earth (9.8 m/s^2)