A cricketer hits a ball with a velocity \(25 \; 111 : 15\) at 60^\circ above the horizontal. How far above the ground it passes over a fielder 50 \(\pi\) from the bat (assume the ball is struck very close to the ground)
Text Solution
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Horizontal component of velocity
\(v_x \quad 25 \cos 60^\circ \quad 12.5\,m/s\)
Vertical component of velocity
\(v_y = 25 \sin 60 = 25 \sqrt{3} \, \mathrm{m/s}\)
Time to cover 50 m distance \(t = \frac{50}{12.5} - 4 \sec\)
The vertical height y is given by
\(y - v_yt - \frac{1}{2}gt^2 - 12.5\sqrt{3} \times 4 - \frac{1}{2} \times 9.8 / 16 - 8.2m\)
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