Question

During a tennis match, a player serves the ball at 27.4 m/s, with the center of the ball leaving the racquet horizontally 2.34 m above the court surface. The net is 12.0 m away and 0.900 m high. When the ball reaches the net, (a) what is the distance between the center of the ball and the top of the net? (b) Suppose that, instead, the ball is served as before but now it leaves the racquet at 5.00° below the horizontal. When the ball reaches the net, what now is the distance between the center of the ball and the top of the net? Enter a positive number if the ball clears the net. If the ball does not clear the net, your answer should be a negative number. Use g=9.81 m/s2.

Answer 1

a. The ball's horizontal and vertical positions at time $t$ are given by

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t$

$y=2.34\,\mathrm m-\dfrac g2t^2$

The ball reaches the net when $x=12.0\,\rm m$:

$12.0\,\mathrm m=\left(27.4\,\dfrac{\rm m}{\rm s}\right)t\implies t=0.438\,\rm s$

At this time, the ball is at an altitude of

$2.34\,\mathrm m-\dfrac g2\left(0.438\,\mathrm s\right)^2=1.40\,\mathrm m$

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

$x=\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t$

$y=2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)t-\dfrac g2t^2$

The ball reaches the net at time $t$ such that

$\left(27.4\,\dfrac{\rm m}{\rm s}\right)\cos(-5.00^\circ)t=12.0\,\mathrm m\implies t=0.440\,\mathrm s$

at which point the ball's vertical position would be

$2.34\,\mathrm m+\left(27.4\,\dfrac{\rm m}{\rm s}\right)\sin(-5.00^\circ)\left(0.440\,\mathrm s\right)-\dfrac g2\left(0.440\,\mathrm s\right)^2=0.343\,\mathrm m$

so that the ball does not clear the net with 0.343 m - 0.900 m = -0.557 m.