Question

A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the water at the bottom of the well 5.00 seconds after being dropped from the tower. Calculate the depth of the well. Given: g = -9.81 meters/second2

Answer 1

time taken to reach the ground level=
$t1 = \sqrt{ \frac{2h}{g} } = \sqrt{ \frac{2 \times 100}{9.81} } = 4.5s$
let time taken by stone to reach the water level from ground level = t2
$t2 = (5 - 4.5)s = 0.5s$
and the speed of stone when ot reaches the ground level is given by
$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 100} = 44.3 \frac{m}{s}$
therefore deptg of well is
$h = ut + \frac{1}{2} g {t }^{2} \\ where \: u = 44.3 \frac{m}{s } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: t = 0.5s \\ h = (44.3 \times 0.5) + \\ (\frac{1}{2} \times 9.81 \times {0.5}^{2} ) = 23.37m$