A ball thrown down the incline strikes at a point on the incline 25m below the horizontal as shown in the figure. If the ball rises to a maximum height of 20m above the point of projection, the angle of projection `alpha` (with horizontal x axis) is

Options

A. `tan^(-1)(4/3)`
B. `tan^(-1)(3/4)`
C. `tan^(-1)(3/2)`
D. `tan^(-1)(2/3)`

Answer

A

Explanation

`20=(u^(2)sin^(2)x)/(2g) implies u sinalpha=20`
`-25=20t-1/210t^(2)implies t=5s`
`75=(ucos alpha)t`
Solve to get `tan alpha=4/3`