Question: A cricketer hits a ball with a velocity $25m/s$ at $60^{o}$ above the horizontal. How far above ...

Question

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

Answer 1

Solution

Horizontal component of velocity

$v_{x} = 25\cos 60{^\circ} = 12.5m/s$

Vertical component of velocity

$v_{y} = 25\sin 60{^\circ} = 12.5\sqrt{3}m/s$

Time to cover 50 m distance $t = \frac{50}{12.5} = 4\sec$

The vertical height y is given by

$y = v_{y}t - \frac{1}{2}gt^{2} = 12.5\sqrt{3} \times 4 - \frac{1}{2} \times 9.8 \times 16 = 8.2m$

Question: A cricketer hits a ball with a velocity \(25m/s\) at \(60^{o}\) above the horizontal. How far above ...

Solution