Answer:
Approximately [tex]122.625\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex], the ball was launched from ground level, and that the drag on the ball is negligible.)
Explanation:
Let [tex]v_{0}[/tex] denote the velocity at which the ball was thrown upward.
If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly [tex](-v_{0})[/tex]. The velocity of the ball would be changed from [tex]v[/tex] to [tex](-v_{0})\![/tex] (such that [tex]\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})[/tex]) within [tex]t = 10\; {\rm s}[/tex].
Also because the drag on the ball is negligible, the acceleration of the ball would be [tex]a = -g = -9.81\; {\rm m\cdot s^{-2}}[/tex]. Thus:
[tex]\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}[/tex].
Since [tex]\Delta v = (-2\, v_{0})[/tex]:
[tex]-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}[/tex].
[tex]\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}[/tex].
The ball reaches maximum height when its velocity is [tex]v_{1} = 0\; {\rm m\cdot s^{-1}}[/tex]. Apply the SUVAT equation [tex]x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)[/tex] to find the displacement [tex]x[/tex] between the original position (ground level, where [tex]v_{0} = 49.05\; {\rm m\cdot s^{-1}}[/tex]) and the max-height position of the ball (where [tex]v_{1} = 0\; {\rm m\cdot s^{-1}}[/tex].)
[tex]\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
