Question

A cricketer hits a ball with a velocity 25 m/s at ${60}^{\circ }$ 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)

• A. 8.2 m
• B. 9.0 m
• C. 11.6 m
• D. 12.7 m

Answer 1

The correct option is A

8.2 m

Horizontal component of velocity
$v_x=25cos{60}^{\circ }=12.5m/s$
Vertical component of velocity
$v_y=25sin{60}^{\circ }=12.5\sqrt{3}m/s$

Time to cover 50 m distance $t=\frac{50}{12.5}=4sec$
The vertical height y is given by
$y=v_{y}t-\frac{1}{2}g{t}^2=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times 16=8.2m$