Question

A hitter during batting practice hits a ball at 167.2 km/h at an angle of 57.2 degrees to
the horizontal. How high does the ball go?

Answer 1

Let's write the equation of motion of the ball in vertical direction:

$$y=v_0tsin\theta-\dfrac{1}{2}gt^2, (1)$$

here, $v_0 = 167.2\ \dfrac{km}{h}$ is the initial velocity of the ball, $t$ is the time of flight of the ball, $\theta=57.2^{\circ}$ is the launch angle, $y$ is the vertical displacement of the ball (or the height)  and $g=9.8\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

Let's first find the time that the ball takes to reach the maximum height from the kinematic equation:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$ into the first equation and find the maximum height:

$$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(167.2\ \dfrac{km}{h}\cdot \dfrac{1000\ m}{1\ km}\cdot \dfrac{1\ h}{3600\ s})^2\cdot sin^257.2^{\circ}}{2\cdot 9.8\ \dfrac{m}{s^2}}=77.76\ m.$$

Answer:

$y_{max}=77.76\ m.$