Q. 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 th the ground)
A) $8.2\ m$
B) $9.0\ m$
C) $11.6\ m$
D) $12.7\ m$
Explanation
(a) Horizontal component of velocity
$\upsilon_{x} = 25\cos{60{^\circ}} = 12.5\ m/s$
Vertical component of velocity
$\upsilon_{y} = 25\sin{60{^\circ}} = 12.5\sqrt{3}\ m/s$
Tim to over $50\ m$ distance $t = \frac{50}{12.5} = 4\ sec$
The vertical height $y$ is given by
$y = \upsilon_{y}t - \frac{1}{2}gt^{2} = 12.5\sqrt{3} \times 4 - \frac{1}{2} \times 9.8 \times 16 = 8.2\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.