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))`

Options

A. `tan^(-1)(1.2)`
B. `tan^(-1)(0.2)`
C. `cot^(-1)(2)`
D. `cot^(-1)(2.5)`

Answer

C

Explanation

`"emf" =e = 3E sin omegat +4E cos omegat =sqrt((3E)^(2)+(4E)^(2))sin(omegat+phi_(1))=5Esin(omegat+phi_(1))` where
`phi_(1)=tan^(-1).(4)/(3)=53^(@)`
Current `i=4|sinomegat-3|cosomegat=sqrt((4|)^(2)+(3|)^(2))sin(omegat-phi_(1))=5lsin(omegat-phi_(2))`
where `phi_(2)=tan^(-1).(3|)/(4|)=37^(@)`
Phase difference `phi=phi_(1)-(-phi_(2))=phi_(1)+phi_(2)=53^(@)+37=90^(@)`
Power `=e_("rms")i_("rms")cosphi=0`