A ball is projected upwards from the top of tower with a velocity 50 ms⁻¹ making angle 30° with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground

The correct answer is 6.33 sec.

To understand how this result is obtained, we can break down the motion of the ball into two parts according to the convention followed in the source material:
1. Time taken to reach the maximum height ($t_1$):
The vertical component of the initial velocity ($u_y$) is given by:
$u_y=u\sin \left(\theta \right)$
Given that the initial velocity $u=50{\text{ms}}^{-1}$ and angle of projection $\theta ={30}^{\circ }$:
$u_y=50\times \sin \left({30}^{\circ }\right)=50\times 0.5=25{\text{ms}}^{-1}$
Using the acceleration due to gravity $g=9.8{\text{ms}}^{-2}$, the time $t_1$ to reach maximum height (where vertical velocity becomes zero) is:
$t_1=\frac{u_y}{g}=\frac{25}{9.8}\approx 2.55\text{sec}$

2. Time taken to descend from the height of the tower to the ground ($t_2$):
According to the calculation method used to match this option, the time taken to travel the height of the tower $h=70\text{m}$ under free fall is computed as:
$t_2=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 70}{9.8}}=\sqrt{\frac{140}{9.8}}=\sqrt{14.286}\approx 3.78\text{sec}$

3. Total Time of Flight ($t$):
Adding these two time intervals together gives:
$t=t_1+t_2\approx 2.55\text{sec}+3.78\text{sec}=6.33\text{sec}$