From the top of a tower of height $40\text{m}$, a ball is projected upwards with a speed of $20\text{m/s}$ at an angle of elevation of ${30}^{\circ }$. Then the ratio of the total time taken by the ball to hit the ground to the time taken by the ball to come at the same level as the top of the tower (Take $g=10{\text{ms}}^{-2})$

The correct option is A $2:1$



Time is taken by the ball to reach the level of the tower $OA$
$T=\frac{2v\sin \theta }{g}=\frac{2v\sin {30}^{\circ }}{g}$
$T=2\text{s}$

Time is taken by the ball to hit the ground after the same level as the top of the tower$\left(AB\right)$
$S=ut+\frac{1}{2}a{t}^2$
$-40=-v\sin {30}^{\circ }t-5t^2$
$40=10t+5t^2$
Solving the above equation we get,
$t=2\text{s},t=-4\text{s}$

Then the total time taken by the ball to hit the ground will be
$T^{\prime }=T+t=2+2=4\text{s}$

Then, the ratio will be
$T^{\prime }:T=4:2$
$=2:1$

Hence, option $\left(\text{A}\right)$ is correct.