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Homework Help: Tennis kinematics problem

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data

    A tennis player hit the ball giving it velocity v0(speed) whose direction is horizontal. Neglecting the air velocity.Are given (in meters) :
    height of player:2,40m
    Distance from player to the net of the tennis court:15m
    Height of the net:0,9m
    What is the smallest value for V0(velocity) which the ball can pass through the net?

    2. Relevant equations
    v=v0+at
    s=so+v0t (UM)
    s=so+vot+at²(1/2)
    v²=vo+2aY
    A= (vo²sen(2theta)/g
    I'm not recalling right now the other...


    Right answer : 27m/s
     
  2. jcsd
  3. Mar 15, 2010 #2
    Re: kinematics

    Think of what criteria does it have to possess in order to pass the net.
     
  4. Mar 15, 2010 #3
    Re: kinematics

    Thank u 4 not helping.
     
  5. Mar 15, 2010 #4
    Re: kinematics

    I am sorry if this wasn't what you expected. Appropriate helping is guiding to the right path, not spoon feeding. At least tell me what you think the answer should be. If it were wrong, I can tell you what went wrong. You have to show that you have tried.
     
  6. Mar 15, 2010 #5
    Re: kinematics

    How about if you list all the known values and decide which equations can be used?
     
  7. Mar 15, 2010 #6
    Re: kinematics

    My friend i do have tried a lot and i won't post it becouse i'm not willing to do it for i did it so many times on paper .I like to answer every question that is in my book of mechanics.This is one of the questions that i am not able yet to find an answer...
     
  8. Mar 15, 2010 #7
    Re: kinematics

    I see. Have you try listing the known values? It won't take long. Plus if you do so, you might see the solution at once; if not, we will know what you have missed.
     
  9. Mar 15, 2010 #8
    Re: kinematics

    I'm sorry english is not my mother-tongue but by known values what exacly you mean? You want me post the given datas in feet mesures?
     
  10. Mar 15, 2010 #9
    Re: kinematics

    ya, the given data. doesn't matter if in feet or metres.
     
  11. Mar 15, 2010 #10
    Re: kinematics

    Do u need glasses?
     
  12. Mar 15, 2010 #11
    Re: kinematics

    Are you trying to challenge my patience? If yes, please note that I am just trying to help. I am not getting paid from this.
    And maybe I should have been more clear. I meant "known variables". e.g. [tex]V_{y0}=0[/tex]
     
  13. Mar 15, 2010 #12
    Re: kinematics

    Sorry I'm just laughing right now,but no.I think Vo is different of zero.I guess it is equal to v²=20Y or V= 10t.
     
  14. Mar 15, 2010 #13
    Re: kinematics

    It is important that you list them out if you don't know how to start, like this:
    Here:
    [tex] x-direction: x_0=0[/tex]
    [tex]x_f=50[/tex]
    [tex]\Delta x=v\Delta t[/tex]

    [tex]v_0= v_f[/tex]
    [tex]\Delta t=?[/tex]

    Now, how about if you list all the y-direction variables?
     
  15. Mar 15, 2010 #14
    Re: kinematics

    Well,lets focus on my book picture:there's a tennis with a net right in the middle 0,9 meters hight and it seems like the movement of the ball is a semi parable.And the question asks for a initial velocity and is not given any other velocity is known that horizontal velocity is constant but i dont know the module and what can i say...
     
  16. Mar 15, 2010 #15
    Re: kinematics

    I think the question has a problem so.
     
  17. Mar 15, 2010 #16
    Re: kinematics

    you need to find the time it takes for the ball to reach the top of the net.
    this is the same time as if it were to fall from 2.4m height, straight down to 0.9m
     
  18. Mar 15, 2010 #17

    ideasrule

    User Avatar
    Homework Helper

    Re: kinematics

    If the ball passes through the net, what do you know about its height at the moment it passes the net? Use that to help you.
     
  19. Mar 16, 2010 #18
    Re: kinematics

    You only need one equation to answer this problem.

    Do you know any calculus? If so that will help you understand everything about this equation & how to derive it.

    In any case, here is the equation;
    [tex]x \ = \ x_0 \ + \ (v_0_x)t \ + \ \frac{1}{2} a t^2 [/tex]

    This equation, used properly, will answer every kinematic question (well, not crazy hard ones or anything :tongue2: ) & derive all of the ones you listed by basic algebra.

    This equation works in two dimensions, horizontal and vertical. Here is it in the vertical (y) direction.

    [tex]y \ = \ y_0 \ + \ (v_0_y)t \ + \ \frac{1}{2} a t^2 [/tex]

    Now, acceleration is downwards so we'll set it to be negative.
    x = final position in the x-axis (this will be 15m)
    x_0 = position. (this will be zero for us here)
    v_0_x = initial position in the x direction, (this is what we want)
    v_0_y = initial velocity in the y direction (this will be zero because the question told us that the ball was hit horizontally).
    a = acceleration (in the x-axis direction there is no acceleration because there is no force to cause it to accelerate, in the y-axis there is only the acceleration of gravity downwards and it is always 9.8 m/s^2 near sea level - use this value).

    Play around with these two equations, some parts of the equations will be zero & wont be used in this example. Try thinking about it, if you solve for some variable in one equation how can you substitute into the other equation in a way to solve for what you want.

    When you get the answer, try to play around with these amazing equations to derive all of the equations you listed in your list.

    I'll give a REALLY proper hint as you're new to this magical equation.

    In the y-axis equation, the final height y is 2.4 and the initial height is 0.9. Go!
     
  20. Mar 16, 2010 #19
    Re: kinematics

    2,4 = Vo*t + 5*t
    Knowing that the initial velocity is as far as i can see nul so:

    t = 0,5s

    Vo = [tex]\frac{15}{0,5}[/tex]

    Vo = 30m/s

    Thats all that i could get.Anyway have you guys already founf the answer for vectorial module of Vo ?Which is 27m/s?
     
  21. Mar 16, 2010 #20
    Re: kinematics

    sorry an error: ...5t²
     
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