A boy throws a ball upwards with velocity $u = 15ms^{- 1}$. The wind imparts a horizontal acceleration of $3ms^{- 1}$to the left. The angel $ heta$ at which the ball must be thrown so that the ball returns to the boy’s hand is $(use\ g = 10\ ms^{- 2})$

Question

A boy throws a ball upwards with velocity $u = 15ms^{- 1}$. The wind imparts a horizontal acceleration of $3ms^{- 1}$to the left. The angel $	heta$ at which the ball must be thrown so that the ball returns to the boy’s hand is $(use\ g = 10\ ms^{- 2})$

Dr. Abubacker · Accepted Answer

(c) Here $u_{y} = u\cos{	heta = 15}\cos	heta$
$u_{x} = u\sin	heta = 15\sin	heta$
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Time of flight of the ball is
$T = \frac{2u_{y}}{g} = \frac{2 	imes 15\cos	heta}{10} = 3\cos	heta$ …(i)
The boy will catch the ball in time $T$, displacement of the ball in horizontal direction should also be zero, so
$0 = u_{x}T - \frac{1}{2}a_{x}T^{2}$
or $T = \frac{2u_{x}}{a_{x}} = \frac{2(15\sin	heta)}{3} = 10\sin	heta$ …(ii)
From Eqs. (i) and (ii), $3\cos{	heta = 10\sin	heta}$
or $	an	heta = \frac{3}{10} = 0.3$ or $	heta = 	an^{- 1}{(0.3)}$

Q. A boy throws a ball upwards with velocity $u = 15ms^{- 1}$. The wind imparts a horizontal acceleration of $3ms^{- 1}$to the left. The angel $\theta$ at which the ball must be thrown so that the ball returns to the boy’s hand is $(use\ g = 10\ ms^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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