# Distance between tennis ball and the top of the net

1. Jun 15, 2008

### goaliejoe35

1. The problem statement, all variables and given/known data

During a tennis match, a player serves the ball at 25.3 m/s, with the center of the ball leaving the racquet horizontally 2.50 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.

I need help trying to set up an equation for this. I am not sure what I should use. could some one please point me in the right direction?

2. Jun 15, 2008

### konthelion

You need to set up two equations, one for the distance in the x direction, and one for the distance in the y direction.
Use the equation: $$x = x_0 + v_0 t + (1/2) a t^2$$ (this can be applied in the y direction as well)

3. Jun 15, 2008

### goaliejoe35

OK so would, x0= 0 and v0= 25.3 m/s? Also would t= 0.474s from the equation t=x/v = t=12/25.3 = 0.474s?

Last edited: Jun 15, 2008
4. Jun 15, 2008

### goaliejoe35

Wait I just confused myself even more. Is x equal to 12m? so plug that in and solve for t?

5. Jun 15, 2008

### konthelion

Ok, now set up the equation for the y-direction. Hint: What is the initial velocity in the y? Also what is the initial height in y? What is the acceleration in this case?

Yeah, you want to find t because that's the time that it takes for it to reach 12m in the x-direction, then you can use that time and plug it back in the y-direction equation to find the height at time t. Then take the difference of that height with the height of the net, .9m

Last edited: Jun 15, 2008
6. Jun 15, 2008

### goaliejoe35

Would the initial velocity in the y be 0?

7. Jun 15, 2008

### konthelion

Yes. Then the initial $$y_{0}$$ would be the height in which the player served the ball, which would be? Then write out the equation and plug back in t.

8. Jun 15, 2008

### goaliejoe35

initial height would be 2.5 m

9. Jun 15, 2008

### goaliejoe35

so wold the equation look like this?

y=2.5+0+.5(9.8)(.437)^2
y=3.44m

then subtract the height of the net from that to get 2.54m?

10. Jun 15, 2008

### konthelion

Ok, then what would $$y = y_0 + v_0 t + (1/2) a t^2$$ be? a=acceleration due to gravity

Close, but no. In what direction is gravity in?

11. Jun 15, 2008

### goaliejoe35

what I did above ^ ??

12. Jun 15, 2008

### goaliejoe35

what do you mean? should it be negative?

13. Jun 15, 2008

### konthelion

Yes, it should since you took the distance 2.5m to be positive, then the direction of gravity is downwards, negative. Then you should get y=? Then the difference between the ball and the net is?

14. Jun 15, 2008

### goaliejoe35

y=2.5+0-.5(9.8)(.437)^2
y=1.56m

1.56-0.9= 0.66m above the net?

15. Jun 15, 2008

### konthelion

Wait, no. Check your time,t, again. It's not correct. Look at your second post

16. Jun 15, 2008

### goaliejoe35

The second post if wrong I think...

12=25.3t+4.9t^2 and solve for t and I get 0.437s

I don't think the way I did it in the second post is quite right.

17. Jun 15, 2008

### konthelion

What? Then how are you getting t?

Dude, you're confusing yourself. Why are you plugging in 12? 12m is the distance in the x direction. You are finding Y. You were doing just fine, you're just mixing up the x and the y.

This is absolutely wrong. There are NO acceleration in the x direction. Reread the entire thread.

Last edited: Jun 15, 2008
18. Jun 15, 2008

### goaliejoe35

from the equation x=x0+v0t+.5gt^2 plug in x=12 v0=25.3 and solve for t

19. Jun 15, 2008

### goaliejoe35

20. Jun 15, 2008

### konthelion

There are two equations
(1) $$y=2.5m-\frac{1}{2}(9.8m/s^2)(t)^2$$
(2) $$x=25.3m/s * t$$

You already solved for t, in (2) when x = 12. Now, plug it back into (1). Then, y would be?