Question

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

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

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:

$T=\frac{2v_y}{g}=\frac{40cos\theta }{10}=4cos\theta (i)$

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, $T=\frac{2v_x}{a_x}=\frac{(20sin\theta {)}^2}{4}$

$=10sin\theta (ii)$

From equations (i) and (ii),

$4cos\theta =10sin\theta$

tan $\theta =\frac{4}{10}=0.4$