Question

A boy throws a ball upward with velocity $v_{o} = 20 m./s.$ The wind imparts a horizontal acceleration of $4 m/s^{2}$ to the left. The angle 6 at which the ball must be thrown so that the ball returns to the boy’s hand is: $(g = 10 m/s^{2})$

• A. $tan^{-1}(1.2)$
• B. $tan^{-1}(0.2)$
• C. $tan^{-1}(2)$
• D. $tan^{-1}(0.4)$

Answer 1

Answer: (D)

$v_{y} =v_{o}$ cos $\theta=20$ cos$\theta$
$v_{x}=v_{o}$sin $\theta=20$ sin $\theta$
Time of flight of the ball is:
$T=\frac{2v_{y}}{g}=\frac{40\,\,cos\theta}{10}=4\,\,cos\theta\,\,\,\,\,\,(i)$
In this time displacement of ball in horizontal direction should also be zero,
i,e., $0=V_{x}T-\frac{1}{2}a_{x}T^{2}$
This gives, $T=\frac{2v_{x}}{a_{x}}=\frac{(20\,\,sin\,\, \theta)^{2}}{4}$
$=10\,\,sin\,\, \theta\,\,\,(ii)$
From equations (i) and (ii),
$4\,\,cos\,\, \theta=10\,\,sin\,\, \theta$
tan $\theta =\frac{4}{10}=0.4$