91Ó°ÊÓ

The initial velocity and the projection angles for the projectile are$v_0$and$\mathrm{θ}$respectively.

The vertical component of the speed $\left(v_y\right)$of the projectile is given by

role="math" $v_y=v_0\sin \mathrm{θ}..................\left(1\right)$

At maximum height the final speed of the projectile is zero.

The formula to calculate the final speed $\left(v\right)$of the projectile is given by

$v^2={v_y}^2-2gh.................\left(2\right)$

Substitute $0$for $v$and $v_0\sin \mathrm{θ}$for $v_y$into equation (2) and solve to calculate the required maximum height.

role="math" localid="1647402886362" $\begin{array}{rcl}0& =& {\left(v_0\sin \mathrm{θ}\right)}^2-2gh\\ h& =& \frac{{v_0}^2{\sin }^2\mathrm{θ}}{2g}....................\left(4\right)\end{array}$

The initial speed of the baseball is$33.6\text{m/s}$and the angles of projections are$30.0°.45.0°,60.0°$.

Substitute the values of the initial speed and the projection angles and $9.80{\text{m/s}}^2$for $g$into equation (4) to calculate the height of the baseball for the given projection angles.

$\begin{array}{rcl}{h}_{30}& =& \frac{{\left(33.6\mathrm{m}/\mathrm{s}\right)}^2×{\sin }^{2}30.0°}{2×9.80{\text{m/s}}^2}\\ & =& 14.4\text{m}\\ h_{45}& =& \frac{{\left(33.6\mathrm{m}/\mathrm{s}\right)}^2×{\sin }^{2}45.0°}{2×9.80{\text{m/s}}^2}\\ & =& 28.8\text{m}\\ h_{60}& =& \frac{{\left(33.6\mathrm{m}/\mathrm{s}\right)}^2×{\sin }^{2}60.0°}{2×9.80{\text{m/s}}^2}\\ & =& 43.2\text{m}\end{array}$

The formula to calculate the distance travelled by the baseball is given by

$x=\frac{{v_0}^2\sin 2\mathrm{θ}}{g}.................\left(5\right)$

Here, $x$is the horizontal distance travelled by the baseball.

Substitute the values of the initial speed and the projection angles and localid="1648261024193" $9.80{\text{m/s}}^2$for $g$into equation (5) to calculate the distance travelled by the baseball for the given projection angles.

localid="1648261097282" $\begin{array}{rcl}{x}_{30}& =& \frac{{\left(33.6\text{m/s}\right)}^2×\sin \left(2×30.0°\right)}{9.80{\text{m/s}}^2}\\ & ≈& 100\text{m}\\ x_{45}& =& \frac{{\left(33.6\text{m/s}\right)}^2×\sin \left(2×45.0°\right)}{9.80{\text{m/s}}^2}\\ & =& 115.2\text{m}\\ x_{60}& =& \frac{{\left(33.6\text{m/s}\right)}^2×\sin \left(2×60.0°\right)}{2×9.80{\text{m/s}}^2}\\ & ≈& 100\text{m}\end{array}$