A boy throws a ball at an angle `theta` with the vertical. If the vertical component of the initial velocity is `20ms^(-1)` and the wind imparts a horizontal acceleration of `8ms^(-2)` to the left, the angle at which the ball must be thrown so that the ball returns to the boy's hand is `theta`. What is the value of `10(tan theta)`

Answer

`08.00`

Explanation

In y direction
`u _(y) = 20 m//s`
`a _(y) =- 10 m//s ^(2)`
`s _(y) = u _(y) t + (1)/(2) a _(y) t ^(2)`
`t = ( 2 u _(y))/( a _(y)) = (2 xx 20)/(10) = 4 sec`
In `alpha` direction
`s _(x) = 0 = mu _(x) t + (1)/(2) a xx d ^(2)`
` 0= u _(x) (4) - (8)/(2) (4) ^(2)`
`16 = 4x `
`tan theta = ( 4x)/(4y) = (16)/(20)`
`10 tan theta = 10 xx (16)/(20) = 8`