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 50m 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
$v_x$ = 25 cos ${60}^{\circ }$ = 12.5 m/s
Vertical component of velocity
$v_y$ = 25 cos ${60}^{\circ }$ = 12.5 $\sqrt{3}$ m/s
Time to cover 50 m distance t = $\frac{50}{12.5}$ = 4 sec

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