Answer:
Approximately [tex]47.9\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance on the baseball is negligible.)
Explanation:
If the air resistance on the baseball is negligible, the baseball will reach maximum height at exactly [tex](1/2)[/tex] the time it is in the air. In this example, that will be [tex]t = (6.25\; {\rm s}) / (2) = 3.125\; {\rm s}[/tex].
When the baseball is at maximum height, the velocity of the baseball will be [tex]0[/tex]. Let [tex]v_{f}[/tex] denote the velocity of the baseball after a period of [tex]t[/tex]. After [tex]t = 3.125\; {\rm s}[/tex], the baseball would reach maximum height with a velocity of [tex]v_{f} = 0\; {\rm m\cdot s^{-1}}[/tex].
Since air resistance is negligible, the acceleration on the baseball will be constantly [tex]a = (-g) = (-9.81\; {\rm m\cdot s^{-2}})[/tex].
Let [tex]v_{i}[/tex] denote the initial velocity of this baseball. The SUVAT equation [tex]v_{f} = v_{i} + a\, t[/tex] relates these quantities. Rearrange this equation and solve for initial velocity [tex]v_{i}[/tex]:
[tex]\begin{aligned}v_{i} &= v_{f} - a\, t \\ &= (0\; {\rm m\cdot s^{-1}}) - (-9.81\; {\rm m\cdot s^{-2}})\, (3.125\; {\rm s}) \\ &\approx 30.656\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The displacement of an object is the change in the position. Let [tex]x[/tex] denote the displacement of the baseball when its velocity changed from [tex]v_{i} = 0\; {\rm m\cdot s^{-1}}[/tex] (at starting point) to [tex]v_{t} \approx 30.656\; {\rm m\cdot s^{-1}}[/tex] (at max height) in [tex]t = 3.125\; {\rm s}[/tex]. Apply the equation [tex]x = (1/2)\, (v_{i} + v_{t}) \, t[/tex] to find the displacement of this baseball:
[tex]\begin{aligned}x &= \frac{1}{2}\, (v_{i} + v_{t})\, t \\ &\approx \frac{1}{2}\, (0\; {\rm m\cdot s^{-1}} + 30.565\; {\rm m\cdot s^{-1}})\, (3.125\; {\rm s}) \\ &\approx 47.9\; {\rm m}\end{aligned}[/tex].
In other words, the position of the baseball changed by approximately [tex]47.9\; {\rm m}[/tex] from the starting point to the position where the baseball reached maximum height. Hence, the maximum height of this baseball would be approximately [tex]47.9\; {\rm m}\![/tex].
