Question

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

Answer 1

`d = u * t \: + (1)/(2) * a * {t}^(2)`
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2


`d = 0.5 * 9.8 * {4.2}^(2)`
on solving we get

d = 86.436 metres


Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2