Question

a cricketer hits a ball with velocity 25m/s at 60 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) 9m c) 11.6m d) 12.7m​

Answer 1

❍ Horizontal component of velocity :

$\sf \qquad\dashrightarrow v_{x} = 25 \: cos60 \degree$

$\sf \qquad \dashrightarrow v_{x} = 25 \: (0.5)$

$\red{ \pmb{\sf \qquad \dashrightarrow v_{x} = 12.5 \: m {s}^{ - 1} }}$

$\:$

❍ Vertical component of velocity :

$\sf \qquad \dashrightarrow v_{y} = 25 \: sin60 \degree$

$\sf \qquad \dashrightarrow v_{y} = 25 \: (\dfrac{ \sqrt{3} }{2} )$

$\red{ \pmb{ \qquad \sf \dashrightarrow v_{y} = 12.5 \sqrt{3} \: m {s}^{ - 1} } }$

$\:$

❍ Time to cover 50m distance is t.

$\sf \qquad \dashrightarrow t = \dfrac{d}{ v_{x} }$

$\qquad \sf \dashrightarrow t = \dfrac{50}{12.5}$

$\red{ \pmb{ \qquad \sf \dashrightarrow t = 4 \: secs}}$

$\:$

❍ The vertical height ‘y’ is given by :

$\sf \qquad \dashrightarrow y = v_{y}t - \dfrac{1}{2} \: g {t}^{2}$

$\sf \qquad \dashrightarrow y = (12.5 \sqrt{3} ) \: (4) \: - \dfrac{1}{2} \: (9.8) \: (16)$

$\sf \qquad \dashrightarrow y = (86.6) \: - \dfrac{1}{2} \: (156.8)$

$\qquad \dashrightarrow \sf y = 86.6 - 78.4$

$\red{ \pmb{ \sf\qquad \dashrightarrow y = 8.2 \: m \: \: \: \: \{0ption \: a \}{}}}$