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
Show Answer
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 $
