Question

A stone is thrown upwards from the top of a tower with some initial speed and it reaches the ground in $16\text{sec}$. Now it is thrown with the same initial speed downward and it reaches the ground in $9\text{sec}$⁡. In how much time (in 𝑠𝑒𝑐𝑜𝑛𝑑) will it reach the ground if the stone is allowed to fall freely under gravity from the same place?

Answer 1

Suppose the body be projected vertically upwards from 𝐴 with a speed $u_0$.
Using equation $s=ut+\frac{1}{2}a{t}^2$
For first case,
$-h=u_0{t}_1-\frac{1}{2}g{t}_1^2\dots \left(i\right)$
For second case,
$-h=-u_0{t}_2-\frac{1}{2}g{t}_2^2\dots \left(ii\right)$
Now, $\left(i\right)$-$\left(ii\right)$, then we get,
$0=u_0(t_2+t_1)+\frac{1}{2}g(t_2^2-t_1^2)$
$u_0=\frac{1}{2}g(t_1-t_2)\dots \left(iii\right)$

Put the value of $u_0$ in $\left(ii\right)$,
$-h=-\frac{1}{2}g(t_1-t_2)t_2-\left(1/2\right)g{t}_2^2$

⇒$h=\frac{1}{2}g{t}_1{t}_2\dots \left(iv\right)$

For third case, $u=0$,$t=?$

$-h=0\times t-\frac{1}{2}g{t}^2$
or $h=\frac{1}{2}g{t}^2\dots \left(v\right)$

Combining equation $\left(iv\right)$ and equation $\left(v\right)$, we get

$\frac{1}{2}g{t}^2=\frac{1}{2}g{t}_1{t}_2$ or
$t=\sqrt{t_1{t}_2}$

Putting the value we get, $t=\sqrt{16\times 9}$
$t=12sec$

Final Answer: 12