# Tennis player hits the ball velocity help

1. Oct 24, 2005

### Chocolaty

I answered the questions and I think i have them right, this is new stuff though so I just wanna make sure i understand everything, if someone could confirm it would put a smile on my face like this ->

During a serve, a tennis player hits the ball horizontally with a force of 500 N, without putting a spin on it. The 100g ball leaves the racket at a velocity of 40 m/s

a) calculate the time during which the ball was in contact with the player's racket

So i use the impulse formula:
Force * Time = Mass * delta velocity
500 * t = 0.1(kg) * 40
t = 0.008s

b) During its displacement, the ball slows down due to the effect of air resistance. Assuming that the frictional force of 5 N is constant, calculate the ball's average horizontal deceleration in the first second of its flight.

Here i'll find the difference between the initial acceleration and the one after 1 second:
Force = Mass * Acceleration
500 = 0.1 * a
a = 5000m/s^2

Fnet = Fa - Ff = ma
500 - 5 = 0.1 * a
a = 4950m/s^2

So the horizontal deceleration is -50m/s^2

c) Calculate the ball's horizontal velocity when it hits the ground, on the other side of the net, at a distance of 12m from the server.

V2^2 - 1600 = 2 * -50 * 12
V2 = sqr(400) = 20m/s

2. Oct 24, 2005

### Chi Meson

a) correct

b) NO NO NO.

The question is asking for the acceleration during the first second of flight. This implies after it has left the racket. The 500 N has done it's work. Now, only the air resistance (and gravity) is on the ball. The question is assuming (erroneously, I think) that during the first second, there will be negligent vertical motion (or are you doing calculus based physics?), so the horizontal accelration will be due to the air resistance force alone.(net force is 5 N).

you did get the right answer, but not for the right reason.

c) this is correct, but notice that (delta v)^2 is NOT the same as (vf^2 - vi^2).