Question

A stone dropped from the top of the building takes 5 seconds to reach the ground. If it is stopped momentarily 3 s after it is dropped and then released again, the average velocity of the body is _____${ms}^{-1}(takeg=10{ms}^{-2})$

• A. $25$
• B. $41.66$
• C. $62.5$
• D. $17.85$

Answer 1

The correct option is D $17.85$

Initial velocity $u=0$, Let height of building is $h$,

Using $s=ut+\frac{1}{2}a{t}^2$, $-h=0\times 5-\frac{1}{2}\times g\times 5^2$ $\Rightarrow h=125m$,

Distance travelled by stone in 3 seconds is $h_1$,

$\Rightarrow -h_1=0\times 3-\frac{1}{2}g\times 3^2$ $\Rightarrow h_1=45m$,

Now stone is stopped and then released, stone has to be travelled remaining distance in time $t$.

Remaining distance$=125-45=80m$

Velocity at this moment $u=0$,

$\Rightarrow -80=0\times t-\frac{1}{2}g{t}^2$ $\Rightarrow t=4s$

$\therefore$ Total time travelled$=3+4=7s$

$\text{Average Velocity}=\frac{\text{Total Displacement}}{\text{Total Time}}$

$\therefore$ Average velocity$=\frac{125}{7}=17.85m{s}^{-1}$