Question

A ball is thrown at some angle with the vertical such that its vertical component of the initial velocity is $20\, ms^{- 1}$ . The wind is blowing horizontally such that it imparts $8\, ms^{- 2}$ 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 $10 \tan \left(\right.\Phi\left.\right)=?$

Answer 1

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