Question

You throw a ball straight up, and it reaches a height of20 mabove your hand before falling back down. What was the speed of the ball just after it left your hand?

Short answer

The speed of the ball just after it left at the hand was 19.8 m/s .

Step-by-step solution

Identification of the given data

The given data is listed as follows,

The height of the ball above the hand before falling back down is, h=20 m

Expression for the kinetic energy and the potential energy

The expression for the kinetic energy is as follows,

$\mathit{K}\mathbf{.}\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Here,m is the mass, andvis the speed.

The expression for the potential energy is as follows,

$P.E=mgh$

Here, m is the mass, g is the acceleration due to gravity with value $93.8\mathrm{m}/{\mathrm{s}}^2$, and his the height of the object.

Determination of the speed of the ball

The initial and final energy of the ball is being conserved, so, the expression is as follows,

$$\begin{gathered} E_i=E_f \\ \frac{1}{2}m{v}^2=mgh \\ v=\sqrt{2gh} \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} v=\sqrt{2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(20\mathrm{m}\right)} \\ =19.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the ball just after it left the hand was $19.8\mathrm{m}/\mathrm{s}$