A ball is thrown vertically upwards from the ground. It crosses a point at the height of 25 m twice at an interval of 4 secs. The ball was thrown with the velocity of

Let initial velocity of the ball = u

$\therefore 2S=ut-\frac{1}{2}g{t}^2=ut-5t^2$

$If{t}_{1}and{t}_{2}arethetwosolutions.$

Then $t_2-t_1=\sqrt{{(t_2+t_1)}^2-4t_1{t}_2}$

$\Rightarrow 4^2=\frac{u^2}{2S}-4\times \frac{25}{5}$

$\Rightarrow {16}^2=\frac{u^2}{25}-20\Rightarrow u^2=25\times 36$

$\Rightarrow u=30m/s$