Top tennis players can serve a ball at a velocity of 200km/h (about 125mph) or more. The drag due to the air slows the ball after it leaves the racket. About what fraction of the ball's original velocity will be lost by the time it hits the court?
Assume that the acceleration due to gravity is negligible. The ball starts with a speed of 200km/h and travels 18m before hitting the ground. The density of the air is 1.18kg/m . The mass of the ball is 57g, its diameter is 66mm. The drag coefficient is ; neglect the effect of nap (fuzziness of the ball) and the spin of the ball. Bonus: About how many times is the drag force larger than the force of gravity?
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The force acting on the ball is F = 2 1 c f ρ A v 2 , where c f = 0 . 4 7 is the drag coefficient for a sphere, ρ is the density of the air, A is the cross section of the circular area facing the wind and v is the velocity. The equation of motion for the ball is
d t d v = − 2 m c f ρ A v 2 = − k v 2 ,
where m is the mass, v is the velocity and k = 2 m c f ρ A = 0 . 0 1 6 6 m − 1 . Solving this nonlinear differential equation is not trivial, but it becomes much simpler if we introduce the position x as a variable, instead of the time t . We use d t d v = d t d x d x d v = v d x d v and we get
v d x d v = − k v 2 or d x d v = − k v
At x = 0 the velocity is v = v 0 . With this initial condition the solution is v = v 0 e − k x or v / v 0 = e − k x . We are interested in the quantity ( v 0 − v ) / v 0 = 1 − e − k x = 0 . 2 5 9 . The loss of velocity is about 25%.
Bonus: Initially the drag force on the ball is F = 2 1 c f ρ A v 2 = 2 . 9 N . The weight is m g = 0 . 5 6 N that is more than 5 times less than the drag force. By the time the ball reaches the ground the drag force is only about 2 times larger than the weight.
Here is a link to a serious experimental study . According to the study the drag coefficient is actually a bit larger, c f = 0 . 5 , probably due to the fuzziness of the ball.