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

Given:
Initial velocity,$\mathrm{u}=\mathrm{ }0 \mathrm{m}/\mathrm{s}$
Distance traveled, $\mathrm{s}=\mathrm{ }20 \mathrm{m}$
Acceleration,$\mathrm{a}=10 \mathrm{m}/{\mathrm{s}}^2$
Final velocity,$\mathrm{ }=\mathrm{v}$
Time taken, $\mathrm{t} =\mathrm{ }?$
From the third equation of motion 
$v^2= u^2+2\mathit{as}$
   $= 0^2+2\times 10\times 20$
     $=400$
$\Rightarrow {\mathrm{v}}^2=\mathrm{ }400$
 $\Rightarrow \mathrm{v}=\mathrm{ }\sqrt{400}$
 $\Rightarrow \mathrm{v}=20 \mathrm{m}/\mathrm{s}$
Now, from the first  equation of motion
$\mathrm{v} = \mathrm{u} + \mathrm{at}$        

$\Rightarrow 20=\mathrm{ }0\mathrm{ }\times 10\times \mathrm{t}$
   $\Rightarrow \mathrm{t}=\frac{20}{10}$
Or,   $\mathrm{t} =2 \mathrm{s}$
So, the ball will strike the ground with velocity 20 $\mathrm{m}/\mathrm{s}$  after 2 seconds.