Question

A stone dropped from a certain height reaches ground in $5\text{sec}$. If it is dropped and stopped for infinitesimal time interval at $t=3\text{s}$, then time taken by the stone to reach the ground is

• A. $6\text{s}$
• B. $6.5\text{s}$
• C. $7\text{s}$
• D. $7.5\text{s}$

Answer 1

The correct option is C $7\text{s}$


Given:
Under free fall, ball will reach ground in $5\text{s}$.
(Taking downward motion of ball as $+ve$)
$s=ut+\frac{1}{2}a{t}^2$
$h=0+\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 5^2=125\text{m}$

From $t=0$ to $t=3\text{s}$.

Using,
$s=ut+\frac{1}{2}a{t}^2$
$h_1=0+\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 3^2=45\text{m}$

At $t=3\text{s}$ ball is stoped for infinitesimal time interval which again makes the velocity of ball $v^{\prime }=0\text{m/s}$.
As time interval is infinitesimal small,
$\therefore$ we will neglect it while finding total time of fall.


$h_2=h-h_1=(125-45)\text{m}=80\text{m}$
Again using,
$s=u{t}_2+\frac{1}{2}a{t}_2^2$
$h_2=0+\frac{1}{2}g{t}_2^2\Rightarrow t_2=\sqrt{\frac{2h_2}{g}}\text{s}=4\text{s}$
$\therefore$Total time of fall=$(3+4)\text{s}=7\text{s}$