A cricketer hits a ball with a velocity 25 m/s at 60^o above the horizontal. How far above the ground it passes over a fielder 50m from the bat (assume the ball is struck very close to the ground)

The horizontal component of velocity is given by

${\mathrm{V}}_{\mathrm{X}}=25\cos {60}^{\circ }=12\cdot 5\mathrm{m}/\mathrm{s}$

Similarly, the vertical component of velocity is given by

${\mathrm{v}}_{\mathrm{y}}=25\sin {60}^{\circ }=12\cdot 5\times \sqrt{3}\mathrm{m}/\mathrm{s}$

Time to cover 50 metre distance 

$\mathrm{t}=\frac{50}{12\cdot 5}=4\sec$

The vertical height x is given by (see fig. 10) 

$\begin{array}{l}\mathrm{x}={\mathrm{V}}_{\mathrm{x}}\mathrm{t}-\frac{1}{2}\overline{\mathrm{g}}{\mathrm{t}}^2\\ =12\cdot 5\sqrt{\overline{3}}\times 4-\frac{1}{2}\times 9\cdot 8\times 16\\ =8\cdot 2\mathrm{m}\end{array}$