A cricketer hits a ball with a velocity $25\,\,m/s$ at ${60^o}$ above the horizontal. How far above the ground it passes over a fielder $50 m$ from the bat ........ $m$ (assume the ball is struck very close to the ground)
A cricketer hits a ball with a velocity $25\,\,m/s$ at ${60^o}$ above the horizontal. How far above the ground it passes over a fielder $50 m$ from the bat ........ $m$ (assume the ball is struck very close to the ground)
$8.2 $
$9.0 $
$11.6$
$12.7 $
A cricketer hits a ball with a velocity $25\,\,m/s$ at ${60^o}$ above the horizontal. How far above the ground it passes over a fielder $50 m$ from the bat ........ $m$ (assume the ball is struck very close to the ground)
Horizontal component of velocity
${v_x} = 25\cos 60^\circ = 12.5\,m/s$
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
$y = {v_y}t - \frac{1}{2}g{t^2} = 12.5\sqrt 3 \times 4 - \frac{1}{2} \times 9.8 \times 16 = 8.2\,m$
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