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} $



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