Question

A boy throws a ball upwards with velocity $v_0$ = 20 m/s . The wind imparts a horizontal acceleration of 4 m/$s^2$ to the left.

The angle $\theta$ at which the ball must be thrown so that the ball returns to the boy's hand is ( g = 10m/$s^2$ )

• A. $ta{n}^{-1}\left(\frac{1}{2}\right)$
• B. $ta{n}^{-1}\left(0.2\right)$
• C. $ta{n}^{-1}\left(2\right)$
• D. $ta{n}^{-1}\left(0.4\right)$

Answer 1

The correct option is D

$ta{n}^{-1}\left(0.4\right)$

(d) $v_y=v_{0}cos\theta =20cos\theta$

$v_x=v_{0}sin\theta =20sin\theta$

Time of flight of the ball is :

T = $\frac{2v_y}{g}=\frac{2\times 20cos\theta }{10}=4cos\theta$ ............... (1)

In this time displacement of ball in horizontals Directions should also be zero.

i.e. 0 = $v_{x}T-\frac{1}{2}{a}_x{T}^2$ ; this gives

T = $\frac{2v_x}{a_x}=\frac{2\left(20sin\theta \right)}{4}$ = 10 sin $\theta$ ............. (2)

Equating (1) and (2) we get , 4 cos $\theta$ = 10 sin $\theta$ or tan $\theta$ = $\frac{2}{5}$ ; or $\theta$ = $ta{n}^{-1}$ (0.4)