A cricketer hits a ball with a velocity 25 m/s at 60o 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 byVX=25cos⁡60∘=12⋅5m/sSimilarly, the vertical component of velocity is given byvy=25sin⁡60∘=12⋅5×3m/sTime to cover 50 metre distance t=5012⋅5=4secThe vertical height x is given by (see fig. 10) x=Vxt−12g¯t2=12⋅53¯×4−12×9⋅8×16=8⋅2m