Question

A boy throws a ball upwards with velocity $v_0=20{ms}^{-1}$. The wind imparts a horizontal acceleration of $4{ms}^{-2}$ to the left. The angle $\theta$ at which the ball must be thrown so that the ball returns to the boy's hand is ($g=10{ms}^{-2}$)

• A. ${\tan }^{-1}\left(1.2\right)$
• B. ${\tan }^{-1}\left(2.5\right)$
• C. ${\cot }^{-1}\left(2\right)$
• D. ${\cot }^{-1}\left(2.5\right)$

Answer 1

The correct option is B ${\tan }^{-1}\left(2.5\right)$

$\text{Step 1: Resolving}$ $V_0$

$\text{[Ref. Fig]}$

$V_x=V_{o}sin\theta ;V_y=V_{o}cos\theta$

$\text{Step 2: Time of flight}$

Time of flight $T=\frac{2V_y}{g}=\frac{2V_0\cos \theta }{g}$

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

$\text{Step 3: Angle calculation}$

As the ball returns to the boy's hand.

$\therefore$ displacement in $x$ direction should be zero.

Since acceleration is constant, therefore applying equation of motion in $x$ direction

$x=V_{x}T+\frac{1}{2}{a}_x{T}^2$

From diagram and equation $\left(1\right)$

$0=20\sin \theta \left(4\cos \theta \right)-\frac{1}{2}4(4\cos \theta {)}^2$

$\Rightarrow$ $0=160\sin \theta -64\cos \theta$

$\Rightarrow$ $\cot \theta =2.5$

$\Rightarrow$ $\theta ={\cot }^{-1}\left(2.5\right)$

Hence angle $\theta$ is ${\cot }^{-1}\left(2.5\right)$. Option $D$ is correct.