A baseball is hit 3 feet above the ground at 80 feet per second and at an angle of $45 ^ { \circ }$ with respect to the ground. Find the maximum height. Round your answer to the nearest integer.

Question

Quizplus · Accepted Answer

First, we can use the equation for the height of a projectile motion: 
$$y(t) = v_{0y}t - \frac{1}{2}gt^2$$
where $v_{0y}$ is the initial vertical velocity, $g$ is the acceleration due to gravity, and $t$ is time.

We can find $v_{0y}$ using the initial velocity and the angle: 
$$v_{0y} = v_0 \sin 	heta = (80 	ext{ ft/s})\sin 45^\circ = 56.6 	ext{ ft/s}$$

At the highest point, the vertical velocity will be 0, so we can use this to find the time it takes to reach that point: 
$$0 = v_{0y} - gt_	ext{max} \quad \Rightarrow \quad t_	ext{max} = \frac{v_{0y}}{g} = \frac{56.6 	ext{ ft/s}}{32.2 	ext{ ft/s}^2} \approx 1.76 	ext{ s}$$

We can plug this into the equation for the height to find the maximum height: 
$$y_	ext{max} = v_{0y}t_	ext{max} - \frac{1}{2}gt_	ext{max}^2 \approx 52.9 	ext{ ft}$$

Rounding to the nearest integer, we get $\boxed{	extbf{(C) }53 	ext{ feet}}$.

A Baseball Is Hit 3 Feet Above the Ground at 80