A ball is thrown with a velocity of $25 \mathrm{m}/\mathrm{s}$ in the vertically upward direction. What is the total time taken to return to the surface of the earth. Take  $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$

Given,

$\mathrm{u}=25 \mathrm{m}/\mathrm{s}, \mathrm{v}=0 \mathrm{m}/\mathrm{s}, \mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2$

Let time taken by the ball to reach the top be $\mathrm{t}$.

By first equation of motion

$\mathrm{v}=\mathrm{u}+\mathrm{gt}$

$0=25-10\mathrm{t}$

$10\mathrm{t}=25$

$\Rightarrow \mathrm{t}=\frac{25}{10}$

$\Rightarrow \mathrm{t}=2.5\mathrm{seconds}$

The ball will take $2.5 \mathrm{seconds}$ to reach the top. it will take same time to reach the bottom from the top.

Hence, total time $=2.5+2.5 =5 \mathrm{seconds}$