A ball is thrown vertically upwards. It returns $6s$ later. Calculate:$\left(i\right)$ the greatest height reached by the ball, and $\left(ii\right)$ the initial velocity of the ball. (Take $g=10m{s}^{-2}$)

The correct option is B (i) $45$ $m$, (ii) $30$ $m$ $s^{-1}$

The ball is thrown up and it returns in $6sec$,

Time of accent = Time of decent

So, time to reach the highest point $=6/2=3s$

By 2nd equation of motion

$s=ut+\frac{1}{2}g{t}^2$

To calculate height, consider motion of the ball from highest point to the ground

$H=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 3^2=45m$

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

$0=ut-\frac{1}{2}g{t}^2$

$u=10\times 6/2=30m/s$