## Fundamentals of Physics Extended (10th Edition)

$0.866m$
This time, the ball travels at a $5^{0}$ angle initially. So the time till the ball reaches the net will be $t=\dfrac {L}{v_{0}\cos \alpha }\approx 0.5104s$ where L is the distance from net to ball when ball is served and $α$ is the angle ball served relative to horizontal line. So the height of the ball when it reaches the net will be: $H_{1}=H_{0}+v_{0}t\sin \alpha -\dfrac {gt^{2}}{2}\approx 0.034m$ So $H_{1} < H_{net}$ and the ball will not clear the net. So the distance form the top of the net to the center of the ball will be $x=H_{net}-H_{1}\approx 0.9-0.034\approx 0.866m$