From the top of tower of height 40 m a ball is projected upwards with a speed of 20 m/sec at an angle of elevation of `30^(@).` Then the ration of the total time taken by the ball to hit the ground to its time of flight ( time taken to come back to the same elevation) is ( take `g = 10 m//sec ^(2))`

Options

A. `2:1`
B. `3:1`
C. `3:2`
D. `1:5:1`

Answer

A

Explanation

From `s_(y) = u_(y)t +(1)/(2)a_(y)t^(2)` we have,
`- 40 = (20 sin 30^(@)) t - 5t^(2)` or `t^(2) - 2t - 8 = 0`
Solving this, we get `t = 4s`
`T = (2u sin theta)/(g) = (2xx 20 xx sin 30^(@))/(10) = 2s rArr (t)/(T) = 2`