From the top of a tower of height 40m, a ball is projected upward with a speed of `20ms^(-1)` at an angle of elevation of `30^(@)`. Then the ratio of the total time taken by the ball to hit the ground to the time taken to ball come at same level as top of tower.

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`