Physics Two Dimensional Motion Question 241
Question: 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 (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 =12.5,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}gt^{2}=12.5\sqrt{3}\times 4-\frac{1}{2}\times 9.8\times 16=8.2,m $
BETA
