Answer to Question #290378 in Physics for Dre

Question

A rock falls from rest a vertical distance of 1.32 meter to the surface of a planet in 0.63 second. The magnitude of the acceleration due to gravity on the planet is approximately

Answer 1

Given:

$h=1.32\:\rm m$

$t=0.63\: \rm s$

$g-?$

The time of free falling

$t=\sqrt{2h/g}$

Hence, the acceleration due to the gravity

$g=\frac{2h}{t^2}=\rm\frac{2*1.32\: m}{(0.63\: s)^2}=6.65\: m/s^2$