A baseball is seen to pass upward by a window with a vertical speed of 14 m/s. If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude did it reach, (c) when was it thrown, and (d) when does it reach the street again?

In the problems related to the description of motion, the kinematic equations are utilized to calculate the speed, time, distance, and other physical quantities.

The kinematic equations used in this problem are as follows.

$v^2=u^2+2gh$.

$v=u+gt$.

Given data.

The vertical speed is $v=14 \mathrm{m}/\mathrm{s}$.

The height below the ground is $h=18\mathrm{m}$.

The relation from the equation of motion is given by

$v^2=u^2+2gh$.

Here, u is the initial speed, and g is the gravitational acceleration.

On plugging the values in the above relation,

$\begin{array}{c}{\left(14 \mathrm{m}/\mathrm{s}\right)}^2=u^2+2\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)\left(18\mathrm{m}\right)\\ u=23.4\mathrm{m}/\mathrm{s}\end{array}$

Thus, $u=23.4\mathrm{m}/\mathrm{s}$is the required initial speed.

The relation from the equation of motion is given by

$v{\text{'}}^2=u^2+2gh\text{'}$.

Here, v’ is the speed at the maximum height, the value of which is zero, and h’ is the required altitude.

On plugging the values in the above relation,

$\begin{array}{c}{\left(0\right)}^2={\left(23.4 \mathrm{m}/\mathrm{s}\right)}^2+2\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)h\text{'}\\ h\text{'}=27.9\mathrm{m}\end{array}$

Thus, $h\text{'}=27.9\mathrm{m}$is the required altitude.

The relation from the equation of motion is given by

$v=u+gt$.

Here, t is the required time.

On plugging the values in the above relation,

$\begin{array}{c}14\mathrm{m}/\mathrm{s}=\left(23.4 \mathrm{m}/\mathrm{s}\right)+\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)t\\ t=0.95\mathrm{s}\end{array}$

Thus, $t=0.95\mathrm{s}$is the required time.

The relation from the equation of motion is given by

$v\text{'}=u+gt\text{'}$.

Here, t’ is the time.

On plugging the values in the above relation,

$\begin{array}{c}0=\left(23.4 \mathrm{m}/\mathrm{s}\right)+\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)t\text{'}\\ t\text{'}=2.38\mathrm{s}\end{array}$

The total time required to reach the street again is calculated as

$\begin{array}{l}T=2t\text{'}\\ T=2\left(2.38\mathrm{s}\right)\\ T=4.76\mathrm{s}\end{array}$

Thus, $T=4.76\mathrm{s}$is the required time.