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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
    s=so+v0t (UM)
    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:
    [tex] x-direction: x_0=0[/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


    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
  22. Mar 16, 2010 #21
    Re: kinematics

    Yes I've calculated the answer on a scrap of paper & got 27 m/s.
    I've given you the exact method to use with or without vectors.
    I've given you more hints than I should have so I can only assume you haven't learned the concepts properly.

    Not to fear, if you'd like to actually learn the method properly & easily just watch all of this lecture & maybe the first 20 minutes of the second lecture in the playlist.

    This lecture will show you how to use the equation & will explain the 10-20 pages in your book that list the 5-6 equations that seem very disconnected.

    Before I seen this video I was going crazy trying to memorize formulas and where to apply them etc... I hope this helps you.

    If you apply yourself & are still confused after watching this video I'll show you step by step how to get the answer.
    Last edited by a moderator: Sep 25, 2014
  23. Mar 16, 2010 #22
    Re: kinematics

    Ok.I just read in your instruction what i thought was enough but this because i read it too fast and didn't realize it was a intruction.Now i'm gonna stop and reflect on what you wrote.
  24. Mar 16, 2010 #23
    Re: kinematics

    Cool, I do advise the video though if you're unsure it's really helpful.
  25. Mar 16, 2010 #24
    Re: kinematics

    Yes,i'm realy not getting it anyway...I think i'm not thinking correctly I knew it from the begining that was necessary see the solution,once i have never done any question like this one before.
    I give up by now.
  26. Mar 16, 2010 #25
    Re: kinematics

    1. Because the ball was hit horizontally (x-axis-direction) that mean that there is an initial velocity in the x-direction.

    2. The question asks what is the smallest velocity needed for the ball to pass through the net.
    If you think about it, there will be no vertical initial velocity, (in the y-axis-direction) because the ball was not hit at an angle, it was hit just horizontally, so that means that the initial velocity in the x-direction is what we want to find.

    3. Just like vectors, we analyze each direction of motion seperately because in projectile motion (which is what this is) what happens in one direction is independent of the other.
    It's very important to understand that we can solve one equation for something like time or velocity, something we do not know in either equation, and then swap this unknown thing for other things like position that we do know.

    [tex] x \ = \ x_0 \ + \ (v_0_x)t \ + \ \frac{1}{2} a t^2 [/tex]


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

    are the two seperate equations we'll be using.

    4. We use the x axis direction and see how far we get.
    Remember, there is nothing to cause an acceleration in this direction.
    Also, we choose the initial position x_0 to be zero.

    [tex] x \ = \ x_0 \ + \ (v_0_x)t \ + \ \frac{1}{2} a t^2 [/tex]

    [tex] x \ = \ 0 \ + \ (v_0_x)t \ + \ 0 [/tex]

    [tex] x \ = \ (v_0_x)t [/tex]

    We know the position x away from the net is 15m. v_0 the initial velocity is what we want.
    The only thing we don't know is the time it will take so we'll play with the other equation
    and solve for time.

    If we do this we can come back to this equation & plug it in.

    5. We'll play with the y-axis equation.
    There is an acceleration due to gravity of 9.8 m/s^2 downwards.
    We'll have to change the a = acceleration sign to a minus to account for this

    There is no initial velocity in the y-axis.

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

    [tex] y \ = \ y_0 \ + \ 0 \ - \ \frac{1}{2} a t^2 [/tex]

    In this equation we can set y_0 [the initial height] to be 2.4m which is the height of the player who hit the ball.
    Think about it, if the guy hits the ball horizontally it will move downwards.
    Gravity is pulling it downwards.
    The question tells us that the net has a height of 0.9m so we want the final height to be equal to this or higher. we'll do what the question asked us to do and set y = 0.9m because that is the minimum.

    Solve for time first, then plug in numbers;

    [tex]y \ = \ y_0 \ + \ 0 \ - \ \frac{1}{2} a t^2 [/tex]

    [tex]0.9m \ = \ 2.4m \ + \ 0 \ - \ \frac{1}{2} (9.8 m/s^2) t^2 [/tex]

    [tex]0.9m \ - \ 2.4m \ = \ - \ \frac{1}{2} (9.8 m/s^2) t^2 [/tex]

    [tex]- \ 1.5 m\ = \ - \ \frac{1}{2} (9.8 m/s^2) t^2 [/tex]

    multiply both sides by minus 2 to cancel the minuses & the fraction;

    [tex] 3m \ = \ (9.8 m/s^2) t^2 [/tex]

    [tex] t^2 \ = \ \frac{3m}{(9.8 m/s^2)} [/tex]

    [tex] t^2 \ = \ 0.30612s^2 [/tex]

    [tex] t = \pm \sqrt{0.30612s^2} [/tex]

    [tex] t = \pm 0.5532 [/tex]

    So, we have the time (the positive value because it physically can't be a negative time in this situation).

    We'll now solve for v_0

    [tex] x \ = \ (v_0_x)t [/tex]

    [tex] (v_0_x) \ = \ \frac{x}{t} [/tex]

    [tex] (v_0_x) \ = \ \frac{15m}{0.5532s} [/tex]

    [tex] (v_0_x) \ = \ 27.11 m/s[/tex]

    [tex] (v_0_x) \ \approx \ 27 m/s[/tex]

    It is perfectly find to round off the 27.11 m/s to 27 m/s.

    This may look complicated but it only takes 20 seconds to do once you know it.

    I advise you to watch the video to get a better understanding.

    Also, I advise you to read this carefully & look at everything, even the units & me using symbols not numbers etc...

    Have a goo day :biggrin:
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