A baseball is thrown straight up with initial speed ${\mathit{v}}_{\mathbf{0}}$. If air resistance cannot be ignored, when the ball returns to its initial height its speed is less than ${\mathit{v}}_{\mathbf{0}}$. Explain why, using energy concepts.

The work${\mathit{W}}_{\mathbf{o}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{r}}$ done by the other forces is equal to the kinetic energy plus total potential energy when the forces other than the gravitational and elastic forces work on a particle.

${\mathit{K}}_{\mathbf{1}}\mathbf{+}{\mathit{U}}_{\mathbf{1}}\mathbf{+}{\mathit{W}}_{\mathbf{o}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{r}}\mathbf{=}{\mathit{K}}_{\mathbf{2}}\mathbf{+}{\mathit{U}}_{\mathbf{2}}$

The mass of the ball, acceleration due to gravity, and the height will be the same as the ball returns to the same height. So, the kinetic energy of the ball will be the same.

$U_1=U_2$

Then,

$K_2=K_1+W_{other}$

It is known that$W_{other}=f_{k}S$ and$f_k=m{a}_k$ . Here, the frictional force opposes the motion of the ball. So, the direction of the frictional force is opposite to the direction of motion of the ball. Therefore,$W_{other}=-W_{other}$ .

So,$K_2<K_1$ since$W_{other}<0$ .

It is known that the kinetic energy is dependent on the speed, and it can be seen that kinetic energy is less, so the speed will also be less while returning than it was initially.

Therefore, the speed will be less because air resistance force will oppose the motion of the ball.