Question

A ball is thrown at some angle with the vertical such that its vertical component of the initial velocity is $20m{s}^{-1}$ . The wind is blowing horizontally such that it imparts $8m{s}^{-2}$ horizontal acceleration on the ball. If the angle with vertical at which the ball must be thrown so that the ball returns to the man's hand is $\Phi$ , then what is $10\tan (\Phi )=?$

Answer 1

Answer: 8

In y direction
$u_y=20m/s$
$a_y=-10m/s^2$
$s_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$t=\frac{2u_y}{a_y}=\frac{2\times 20}{10}=4\sec$
In - $\alpha$ direction
$s_x=0=u_{x}t+\frac{1}{2}a\times d^2$
$0=u_x\left(4\right)-\frac{8}{2}{\left(4\right)}^2$
$16=4x$
$\tan \theta =\frac{4x}{4y}=\frac{16}{20}$
$10\tan \theta =10\times \frac{16}{20}=8$