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Q. A cricket ball is hit with a velocity $25\, ms ^{-1}, 60^{\circ}$ 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 $\sqrt{3}=1.7$ and $g =10 \, ms ^{-2}$

Motion in a Plane

Solution:

$ y=x \tan \theta-\frac{1}{2} \frac{g x^2}{u^2 \cos ^2 \theta}$
$y =50 \tan 60^{\circ}-\frac{10 \times 50 \times 50}{2 \times 25 \times 25 \times \cos ^2 60^{\circ}}=5 \,m$