Question

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?
A. $ 19.6J $
B. $ 4.9J $
C. $ 10J $
D. $ 9.8J $

Answer 1

Hint: To answer the provided issue, we will first use the formula for the maximum height, and then we will use the formula for the energy dissipated by air drag acting on the ball throughout the climb, since the ball only reached $ 8.0m $ owing to the air drag, thus we will use the formula of potential energy for the energy loss.

Complete step by step answer:
Maximum height, if there is no air drag will be
 $ H = \dfrac{{{u^2}}}{{2g}} = \dfrac{{\left( {{{14}^2}} \right)}}{{2 \times 9.8}} = 10m $
However, due to air drag, the ball only reaches a height of $ 8m $ .
As a result, there is an energy loss.
 $ = mg\left( {10 - 8} \right) \\
   = 0.5 \times 9.8 \times 2 \\
   = 9.8J $
Therefore, $ 9.8J $ energy is dissipated by air drag acting on the ball during the ascent.
So, the correct option is: D. $ 9.8J $ .

Note:
That equation remains true when the projectile is reasonably close to the ground and the projectile's whole kinetic energy has been converted to potential energy. That isn't always the case with projectiles. This is due to the fact that some of the Kinetic Energy of a projectile is attributable to horizontal velocity, which is not converted to Potential Energy. In addition, the conversion to potential energy occurs only when the projectile reaches its maximum height. If the vertical velocity is used in the equation, the projectile is shot upwards, and friction is omitted, the equation accurately predicts the projectile's maximum height.