Kinematics Question 579
Question: A cricket ball is hit with a velocity $ 25m{{s}^{-1}} $ , $ 60{}^\circ $ above the horizontal. How far above the ground, ball passes over a fielder 50 m from the bat (consider the ball is struck very close to the ground)?
Take $ \sqrt{3}=\text{1}\text{.7 } $
and $ \text{g = 10 m}{{\text{s}}^{-2}} $
Options:
A) 6.8 m
B)7 m
C) 5 m
D)10 m
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ y=x\tan \theta -\frac{1}{2}\frac{\text{g}{{x^{2}}}}{{{u^{2}}{{\cos }^{2}}\theta }} $
$ y=50\tan 60{}^\circ -\frac{10\times 50\times 50}{20\times 25\times 25\times {{\cos }^{2}}60{}^\circ }=5\text{m} $
