a. The ball's horizontal and vertical positions at time [tex]t[/tex] are given by
[tex]x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t[/tex]
[tex]y=2.34\,\mathrm m-\dfrac g2t^2[/tex]
The ball reaches the net when [tex]x=12.0\,\rm m[/tex]:
[tex]12.0\,\mathrm m=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t\implies t=0.438\,\rm s[/tex]
At this time, the ball is at an altitude of
[tex]2.34\,\mathrm m-\dfrac g2\left(0.438\,\mathrm s\right)^2=1.40\,\mathrm m[/tex]
which is 1.40 m - 0.900 m = 0.500 m above the net.
b. The change in angle gives the ball the new position functions
[tex]x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t[/tex]
[tex]y=2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)t-\dfrac g2t^2[/tex]
The ball reaches the net at time [tex]t[/tex] such that
[tex]\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t=12.0\,\mathrm m\implies t=0.440\,\mathrm s[/tex]
at which point the ball's vertical position would be
[tex]2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)\left(0.440\,\mathrm s\right)-\dfrac g2\left(0.440\,\mathrm s\right)^2=0.343\,\mathrm m[/tex]
so that the ball does not clear the net with 0.343 m - 0.900 m = -0.557 m.
