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. `1:2`
C. `4:1`
D. `1:4`

Answer

B

Explanation

From figure,
The time taken by the ball to come
back to the same height is
`t_(1) = (2 u sin theta)/g =(2xx 20 xx sin30^@)/10 = 2 s `
Let `t_2` be the time taken by the ball to reach the ground .
For vertical motion, `y = u sin theta t_(2) -1/2 "gt" _(2)^(2)`
`:. - 40 =(20 sin30^@) t_(2) -1/2 xx 10 xx t_2^2 = 10t_2 -5t_2^2`
or `t_2^2-2t_2 - g= 0`
On solving , we get `t_2 =4` s
`:. t_1/t_2 =2/4 =1/2`