An elevator without a ceiling is ascending with a constant speed of $\mathbf{10}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. A boy on the elevator shoots a ball directly upward, from a height of 2.0 mabove the elevator floor, just as the elevator floor is 28 mabove the ground. The initial speed of the ball with respect to the elevator is 20 m/s. (a) What maximum height above the ground does the ball reach? (b) How long does the ball take to return to the elevator floor?

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,

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

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,

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