A cricketer hits a ball with a velocity 25 m/s  at 600  above the horizontal. How far above the ground it passes over a fielder 50 m from the bat? (Assume the ball is struck very close to the ground)

Horizontal component of velocityvx=25cos600=12.5 m/sVertical component of velocityvy=25sin600=(12.5)3 m/sTime to cover 50 m distancet=5012.5=4 secThe vertical height y  is given by y=vyt−12gt2=(12.5)3×4−12×9.8×16=8.2 m