Tennis serve
A tennis player strikes a ball with a racket during the serve. The ball was at H=2.5m above the ground level and received an initial speed v parallel to the ground’s surface. What must be the minimum speed of this ball for it to pass over the net that is h=0.96m high? The net is at d=11.9m from the serving position. For this speed what will be the distance D from the service position to the point the ball touches the ground?
Solution.
The drawing illustrating the problem is given below.
It is a projectile motion and we can separately analyze the motion in vertical and horizontal directions.
In the vertical direction it is a free fall motion and the proper formula for the ball dropping to the net level is
H – h = (1/2)gt2 (1)
from this formula we find the time of falling to this level
(2)
During that time the ball must move in a horizontal direction by distance d. As horizontal motion is linear motion with constant speed v, we apply the formula
d = vt (3)
from which
v = d/t (3a)
and after substituting time from formula (2) we get
(4)
Substituting the numbers we get
v = 21.23 m/s which is 76.5 km/h.
To find the distance D from this problem, we write the formula (1) with total height H instead of H - h.
H = (1/2)gt2 (5)
from this, time of falling to the ground is
(6)
and the total distance D is
(7)
Substituting the numbers gives
D=15.16m.
If you look at formula (7) it may suggest that distance D does not depend on gravitational acceleration g. This would be very strange and of course such a conclusion is wrong. The dependence of D from g is “hidden” in distance d in the formula (7), because d depends on t as is evident from formula (3) and t in turn depends on gravitational acceleration g – formula (2).