A baseball is thrown from the roof of a $\mathbf{22}\mathbf{.}\mathbf{0}\mathbf{m}$ tall building with an initial velocity of magnitude $\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}$and directed at an angle of $\mathbf{53}\mathbf{.}\mathbf{1}\mathbf{°}$ above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of $\mathbf{53}\mathbf{.}\mathbf{1}\mathbf{°}$below the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed ?

The given data can be listed below as,

• The height of a building is, $h_i=22.0\text{m}$.
• The initial velocity of the ball is, $u=12.0\text{m/s}$.
• Theangle of inclination of the ball’s throw is, $\theta =53.1°$.
• The final potential head of the ball is, $h_f=0$ (Since comes at the ground).

In this question, the concept of energy conservation will be used to determine the speed of the ball before reaching the ground at a different angle of projections.

According to the energy conservation method, the relation of the speed of the ball just before it strikes the ground is expressed as,

$$\begin{gathered} \left[{\left(KE\right)}_i+{\left(PE\right)}_i\right]=\left[{\left(KE\right)}_f+{\left(PE\right)}_f\right] \\ \left[\frac{1}{2}m{\left(u\sin \theta \right)}^2\right]+\left(mg{h}_i\right)=\left[\left(\frac{1}{2}m{v_f}^2\right)+\left(mg{h}_f\right)\right] \\ \left[\frac{1}{2}{\left(u\sin \theta \right)}^2\right]+\left(g{h}_i\right)=\left[\left(\frac{1}{2}{v_f}^2\right)+\left(g{h}_f\right)\right] \\ {v}_f=\sqrt{{\left(u\sin \theta \right)}^2+2g\left(h_i-h_f\right)} \end{gathered}$$

Here, ${\left(KE\right)}_i$is the initial kinetic energy of the ball, ${\left(PE\right)}_i$is initial gravitational potential energy, ${\left(KE\right)}_f$is the final kinetic energy of the ball, ${\left(PE\right)}_f$is the final gravitational potential energy, is the gravitational acceleration whose value is $9.81{\text{m/s}}^{\text{2}}$and $v_f$ is the speed of the ball just before it strikes the ground.

Substitute all the known values in the above equation.

$$\begin{gathered} v_f=\sqrt{{\left\{\left(12.0\text{m/s}\right)\sin 53.1°\right\}}^2+\left\{2\left(9.81{\text{m/s}}^{\text{2}}\right)\left(22.0\text{m}-0m\right)\right\}} \\ =23\text{m/s} \end{gathered}$$

Since the initial velocity of the ball is at an angle $53.1°$ below the horizontal means the value of the angle $\theta$ can be considered as$\theta =-53.1°$.

According to the energy conservation method, the relation of the speed of the ball just before it strikes the ground if the initial velocity is at an angle of $53.1°$below the horizontal is expressed as,

$$\begin{gathered} \left[{\left(KE\right)}_i+{\left(PE\right)}_i\right]=\left[{\left(KE\right)}_f+{\left(PE\right)}_f\right] \\ \left[\frac{1}{2}m{\left(u\sin \theta \right)}^2\right]+\left(mg{h}_i\right)=\left[\left\{\frac{1}{2}m{\left({v_f}^{\text{'}}\right)}^2\right\}+\left(mg{h}_f\right)\right] \\ \left[\frac{1}{2}{\left(u\sin \theta \right)}^2\right]+\left(g{h}_i\right)=\left[\frac{1}{2}{\left({v_f}^{\text{'}}\right)}^2+\left(g{h}_f\right)\right] \\ \left({v_f}^{\text{'}}\right)=\sqrt{{\left(u\sin \theta \right)}^2+2g\left(h_i-h_f\right)} \end{gathered}$$

Here, ${v_f}^{\text{'}}$is the speed of the ball just before it strikes the ground.

Substitute all the known values in the above equation.

$$\begin{gathered} \left({v_f}^{\text{'}}\right)=\sqrt{{\left\{\left(12.0\text{m/s}\right)\sin \left(-53.1°\right)\right\}}^2+\left\{2\left(9.81{\text{m/s}}^{\text{2}}\right)\left(22.0\text{m}-0m\right)\right\}} \\ =23\text{m/s} \end{gathered}$$

If the effects of air resistance are includedthen the final speed will be reduced because air resistance will decrease the final energy of the ball which results in the speed of the ball will be reduced.

Asthe friction is a non-conservative force then our system will become a dissipative system as the total mechanical energy would not be conserved.

Thus, none of the situations (either part (a) or part (b)) would give the higher speed.