91Ó°ÊÓ

It is given that,

$$\begin{gathered} {\mathrm{y}}_0=28+2=30\mathrm{m} \\ {\mathrm{v}}_{0\mathrm{y}}=20\mathrm{m}/\mathrm{s} \end{gathered}$$

The problem is based on the standard kinematic equations that describe the motion of an object with constant acceleration. Using these equations, themaximum height and time can be found.For example, a, v, y and tare variables, which can be found from kinematic equations.

Formula:

The displacement and velocity in kinematic equation are given as,

$$\begin{gathered} y=v_{0y}t+\frac{1}{2}a{t}^2 \\ {v}_f^2=v_i^2+2a\left(y-y_0\right) \end{gathered}$$ (i)

(ii)

Where,y is total displacement,$v_{0y}$ is the y component of initial velocity, $v_0$is the initial velocity and $v_f$ is the final velocity. is time and t is an acceleration.

Here at maximum height $v_f=0\mathrm{m}/\mathrm{s}$. Thus with equation (ii) the maximum height will be,

$$\begin{gathered} 0={30}^2-2×9.81\left(\mathrm{y}-30\right) \\ \mathrm{So}, \\ \mathrm{y}=76\mathrm{m} \end{gathered}$$

Thus maximum height will be, y =76 m

To find the time, use the equation (i),

$∆y=v_{0y}t+\frac{1}{2}a{t}^2$

Since, ball’s displacement relative to floor is,

$∆y=-2.0\mathrm{m}$

Therefore,

$$\begin{gathered} -2=20t-0.5×9.8t^2 \\ 4.9t^2-20t-2=0 \\ \mathrm{Thus},t=4.2\mathrm{s} \end{gathered}$$