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 20 m 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 alpha)/(2g) rArr u sin alpha = 20`….(i)
For total time of light
`-25 = u sin alphat-1/2 "gt"^2 rArr 5t^2 -20t -25=0`
`t^2-4t-5=0 rArr t=(4pm sqrt(16+20))/2` = 5 sec.
In 5 sec, horizontal displacement =75 m .
Now u cos `alpha` x 5 = 75 `rArr` u cos `alpha` =15 ...(ii)
From (i) and (ii)
`tan alpha = 20/15=4/3 rArr alpha =tan^(-1)(4/3)`