Ball dropped from height h takes t second to reach on the surface of earth. Hence, i.e ‌h=0×t+‌
1
2
gt2 ‌t=√‌
2h
g
⇒‌‌t=√‌
2h
g
.......(i) For another planet, mass Mp=100Me where, Me is mass of the earth: Radius of planet, Rp=10Re where, Re is the radius of earth. The value of gravitational acceleration on the surface of planet, gp=‌
GMp
Rp2
‌=‌
G×100Me
(10Re)2
‌=‌
GMe
Re2
=g Since, the value of gravitational acceleration is same for other planet, hence time taken to cover same height on the planet is same as t seconds.