Answer:
At 30 degrees, height = 14.39 m and distance = 49.83 m
At 45 degrees, height = 28.77 m and distance = 57.54 m
At 60 degrees, height = 43.16 m and distance = 49.83 m
Explanation:
Let ∝ be the angle with respect to the horizontal that the baseball is hit with.
The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).
Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)
[tex]vsin(\alpha) + gt = 0[/tex]
[tex]t = \frac{-vsin(\alpha)}{g}[/tex]
So the vertical distance it travels within time t is
[tex]y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}[/tex]
[tex]y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}[/tex]
[tex]y = \frac{-v^2sin^2(\alpha)}{2g}[/tex]
Similarly the horizontal distance it travels within time t is:
[tex]x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}[/tex]
[tex]x = \frac{-v^2sin(2\alpha)}{2g}[/tex]
We can pre-calcualte the quantity [tex]\frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54[/tex]
So [tex]y = 57.54sin^2(\alpha)[/tex]
[tex]x = 57.54sin(2\alpha)[/tex]
From here we can plug-in the angles values of 30, 45 and 60 degrees
At 30 degrees, height = 14.39 m and distance = 49.83 m
At 45 degrees, height = 28.77 m and distance = 57.54 m
At 60 degrees, height = 43.16 m and distance = 49.83 m
