A boy throws a ball upward with velocity `v_(0)=20m//s`. The wind imparts a horizontal acceleration of `4m//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=10ms^(-2))`

Answer

176

Explanation

`v_(y)=v_(0) cos theta =20 cos theta, v_(x)=v_(0) sin theta=20 sin theta T_(f)=(2v_(y))/(g)=(2 xx 20 cos theta)/(10)=4 cos theta`
`" in "t=T_(f)s, s_(x)=0 rArr 0=v_(x)T_(f)-1/2 a_(x)T_(f)^(2)`
[The boy must project the ball opposite to wind, for it to return to his hands.]
`T_(f)=(2v_(x))/(a_(x)) rArr 4 cos theta =(2 xx 20 sin theta)/(4) rArr 4 cos theta=10 sin theta rArr =4/10 or theta=tan^(-1) (4/10)`