A $0.5kg$ ball is thrown up with an initial speed $14m{s}^{-1}$ and reaches a maximum height of $8.0m$. How much energy is dissipated by air drag acting on the ball during the ascent?

The correct option is D

$9.8Joule$

Step 1: Given data and drawing the diagram

Maximum height in the presence of air drag, $H$ $=8.0m$

Mass of the ball, $m$ $=0.5kg$

Initial speed,$u$ $=14m{s}^{-1}$

Step 2: Find the maximum height in the presence of no air drag

Using the third equation of motion, $v^2=u^2+2\left(-g\right)h$ [value of $g$ is negative as the ball is thrown upward]

$0=u^2-2gh$

$h=\frac{u^2}{2g}$

Maximum height in the presence of no air drag, $h=\frac{u^2}{2g}$

$$\begin{gathered} =\frac{{14}^2}{2\times 9.8} \\ =10m \end{gathered}$$

Step 3: Find the loss in energy

Loss in energy $=$ Difference between the potential energies for maximum height and for the height in presence of no air drag

Loss in energy$=mg\left(H-h\right)$

$$\begin{gathered} =0.5\times 9.8\left(10-8\right) \\ =9.8J \end{gathered}$$

Hence, option $\left(D\right)$ is the correct answer.