Question

A man throws a ball vertically upwards with a velocity of $20\mathrm{m}/\mathrm{s}$. After what time will the ball come back to his hands. (Take $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$).

Answer 1

Step 1: Given:

Let the velocity with which the ball is thrown upward be $\mathrm{u}$. We are given that $\mathrm{u}=20\mathrm{m}/\mathrm{s}$.

Let the velocity with which the ball comes back to hands be $\mathrm{v}$. Here, $\mathrm{v}=0\mathrm{m}/\mathrm{s}$.

Step 2: Finding the time taken by the ball to come back:

According to newton's first equation of motion:

$v=u+gt$

where, $u=$the initial velocity of the object

$v=$the final velocity of the object

$t=$ time taken by the object for the whole journey

$g=$ acceleration due to gravity

By the first equation of motion,

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

Putting the values of $v=0m/s,u=20m/sandg=10m/s^2.$

$\mathit{0}\mathit{}m\mathit{/}s\mathit{=}\mathit{20}\mathit{}m\mathit{/}s\mathit{}\mathit{+}\mathit{(}\mathit{-}\mathit{10}\mathit{}m\mathit{/}{s}^{\mathit{2}}\mathit{)}\mathit{}t$

Adding $-20$ to both sides of the equation.

$\mathit{-}\mathit{20}\mathit{=}\mathit{-}\mathit{10}t$

Divide both sides by $\mathit{-}\mathit{10}\mathit{.}$

$$\begin{gathered} \frac{\mathit{-}\mathit{20}}{\mathit{-}\mathit{10}}\mathit{=}\frac{\mathit{-}\mathit{10}\mathit{}\mathit{t}}{\mathit{-}\mathit{10}} \\ t\mathit{=}\mathit{2}\mathit{}s \end{gathered}$$

Hence, the ball will come back to his hands after $\mathit{2}\mathit{}seconds\mathit{.}$.