A stone is thrown vertically upward with a speed of 24.0 m/s. (a) How fast is it moving when it is at a height of 13.0 m? (b) How much time is required to reach this height? (c) Why are there two answers to (b)?

Given data.

The speed of the stone is $u=24 \mathrm{m}/\mathrm{s}$.

The height is $h=13\mathrm{m}$.

Velocity is represented as a physical quantity that indicates the correlation among the variations in the location of a particle in unit time. In the SI system, it is measured in terms of m/s.

The relation from the equation of motion is given by.

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

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

On plugging the values in the above relation, you get.

role="math" localid="1643004429175" $\begin{array}{c}{v}^2={\left(24\mathrm{m}/\mathrm{s}\right)}^2+2\left(-9.81\mathrm{m}/{\mathrm{s}}^2\right)\left(13\mathrm{m}\right)\\ v=17.9\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=17.9\mathrm{m}/\mathrm{s}$is the speed of the stone.

The relation from the equation of motion is given by.

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

On plugging the values in the above relation, you get.

$\begin{array}{c}\left(13 \mathrm{m}\right)=\left(24 \mathrm{m}/\mathrm{s}\right)t+\frac{1}{2}\left(-9.81 \mathrm{m}/{\mathrm{s}}^2\right)t^2\\ 4.905t^2+24t-13=0\end{array}$

On solving the above equation, you get two values of time.

$\begin{array}{c}t=\frac{-24\pm \sqrt{{\left(24\right)}^2-4\left(4.905\right)\left(-13\right)}}{2\left(4.905\right)}\\ t=0.492\mathrm{s},-5.38\mathrm{s}\end{array}$

Thus, $t=0.492\mathrm{s},-5.38\mathrm{s}$are the required values of time.

In part (b), two values of time required to reach the height are obtained, which are $t=0.492\mathrm{s},-5.38\mathrm{s}$. One is observed while moving up, and the other is observed on the way down.