Question

An object is allowed to fall freely from a tower of height $39.2m$; exactly at the same time another stone is thrown from the bottom of the tower in vertically upward direction with a velocity of $19.6m{s}^{-1}$ Calculate when and where these two stones would meet ?

Answer 1

Given that,

Height $h=39.2m$

Velocity $u=19.6m/s$

Now, from equation of motion

Case (I) when stone dropped from the tower fall freely

$s_1=ut+\frac{1}{2}g{t}^2$

$s_1=0+\frac{1}{2}g{t}^2....\left(I\right)$

Now, case (II) when stone projected upward

$s_1=19.6t-\frac{1}{2}g{t}^2....\left(II\right)$

Now, total distance

$s=s_1+s_2$

$s=\frac{1}{2}g{t}^2+19.6t-\frac{1}{2}g{t}^2$

$39.2=19.6t$

$t=2\sec$

Now, put the value of t in equation (I)

$s_1=\frac{1}{2}\times 9.8\times 4$

$s_1=19.6m$

Now, the distance is

$s=s_1+s_2$

$s_2=39.2-19.6$

$s_2=19.6m$

Hence, Both stones meet from ground in air at the distance of $19.6m$ in $2sec$