Kinematics Question 691

Question: A cricketer his 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)

Options:

A) 8.2 m

B) 9.0 m

C) 11.6 m

D) 12.7 m

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Answer:

Correct Answer: A

Solution:

[a]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 it’s 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 $



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