A ball is thrown from the top of a building 45 m high with a speed 20$m{s}^{-1}$   above the horizontal at an angle of  ${30}^0$  
The time taken by the ball to reach the ground is t sec and the speed of ball just before it touches the ground is $v$  m/sec then.(Take g=$10m{s}^{-2}$ )
 

Given  $v=20m{s}^{-1},\theta ={30}^0,H=45m$
As the ball has been projected at an angle of  ${30}^0$ above horizontal, so first of all we need to analyze the velocity horizontally and vertically. This will be useful while using distance-time relation in horizontal and vertical direction.
$$\begin{gathered} v_{xi}=v\cos {30}^0=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m{s}^{-1} \\ {v}_{yi}=u\sin {30}^0=20\times \frac{1}{2}=10m{s}^{-1} \end{gathered}$$

It will be easy for us to use distance-time relation in vertical as it will involve less calculation.
In y-direction-45= $10r-\frac{1}{2}\times g{t}^2\Rightarrow t^2-2t=0$
Which on solving gives $t=1+\sqrt{10}s$ (positive value), (other value is $1-\sqrt{10}s$ , s a negative value of time is not acceptable)
$$\begin{gathered} v_{yf}=10-10\times (1+\sqrt{10})=-10\sqrt{10}m{s}^{-1} \\ {v}_f=\sqrt{v_{yf}^2+v_{xf}^2}=\sqrt{{\left(10\sqrt{10}\right)}^2+{\left(10\sqrt{3}\right)}^2} \\ =10\sqrt{3}m{s}^{-1} \end{gathered}$$