1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Distance between tennis ball and the top of the net

  1. Jun 15, 2008 #1
    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. jcsd
  3. Jun 15, 2008 #2
    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: [tex]x = x_0 + v_0 t + (1/2) a t^2[/tex] (this can be applied in the y direction as well)
  4. Jun 15, 2008 #3
    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
  5. Jun 15, 2008 #4
    Wait I just confused myself even more. Is x equal to 12m? so plug that in and solve for t?
  6. Jun 15, 2008 #5
    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
  7. Jun 15, 2008 #6
    Would the initial velocity in the y be 0?
  8. Jun 15, 2008 #7
    Yes. Then the initial [tex]y_{0}[/tex] would be the height in which the player served the ball, which would be? Then write out the equation and plug back in t.
  9. Jun 15, 2008 #8
    initial height would be 2.5 m
  10. Jun 15, 2008 #9
    so wold the equation look like this?


    then subtract the height of the net from that to get 2.54m?
  11. Jun 15, 2008 #10
    Ok, then what would [tex]y = y_0 + v_0 t + (1/2) a t^2[/tex] be? a=acceleration due to gravity

    Close, but no. In what direction is gravity in?
  12. Jun 15, 2008 #11
    what I did above ^ ??
  13. Jun 15, 2008 #12
    what do you mean? should it be negative?
  14. Jun 15, 2008 #13
    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?
  15. Jun 15, 2008 #14

    1.56-0.9= 0.66m above the net?
  16. Jun 15, 2008 #15
    Wait, no. Check your time,t, again. It's not correct. Look at your second post
  17. Jun 15, 2008 #16
    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.
  18. Jun 15, 2008 #17
    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
  19. Jun 15, 2008 #18
    from the equation x=x0+v0t+.5gt^2 plug in x=12 v0=25.3 and solve for t
  20. Jun 15, 2008 #19
    ohhhh my bad man i screwed up your right
  21. Jun 15, 2008 #20
    There are two equations
    (1) [tex]y=2.5m-\frac{1}{2}(9.8m/s^2)(t)^2 [/tex]
    (2) [tex]x=25.3m/s * t[/tex]

    You already solved for t, in (2) when x = 12. Now, plug it back into (1). Then, y would be?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?