A ball is gently dropped from a height of $20m$. If its velocity increases uniformly at the rate of $10m{s}^{-2}$, with what velocity will it strike the ground? After what time will it strike the ground?

Step 1: Given Data

Initial velocity $u=0$

Distance traveled by the ball $S=20m$

Uniform acceleration $a=10m{s}^{-2}$

Let the final velocity be $v$.

Let the time taken be $t$.

Step 2: Calculate the FInal Velocity

According to Newton's third law of motion

$v^2=u^2+2aS$

$\Rightarrow v^2=0+2\times 10\times 20$

$\Rightarrow v=20m/s$

Step 3: Calculate the Time Taken

According to Newton's first law of motion

$v=u+at$

$\Rightarrow t=\frac{v-u}{a}$

$\Rightarrow t=\frac{20-0}{10}=\frac{20{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{10{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}$

$\Rightarrow t=2s$

Hence, the ball will strike the ground at a velocity $20m/s$ at the time $2s$.