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Estimate the maximum initial velocity

  1. Sep 18, 2011 #1
    Hi I need some help with this question:

    Estimate the maximum initial velocity you can achieve with a tennis ball

    I thought of using the maximum range formula: R=v2/g but doesn't that imply to the fact that I'm throwing at an angle of 45 degrees?

    Please help me, I really want to solve this problem, thanks in advance!
     
  2. jcsd
  3. Sep 18, 2011 #2

    NascentOxygen

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    Re: Motion

    Throw the ball so it leaves your hand directed horizontally. Gravity will pull it downwards as it travels away. Write down the equations of motion that might be useful in this exercise.
     
  4. Sep 18, 2011 #3
    Re: Motion

    Hi, thanks for the reply, these are the equations of motion:

    11htif7.png

    How do I know which one to use and how to I motivate my answer?

    u = initial velocity
    V= final velocity

    Thanks in advance!
     

    Attached Files:

  5. Sep 18, 2011 #4

    NascentOxygen

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    Re: Motion

    (3) and (4) look useful.
    Apply (3) vertically: you know s (the height above ground where your hand releases the ball), you know u, you know a. The only unknown is t.
    Apply (4) horizontally.

    Solve the equations simultaneously to determine t, the time that the ball was in flight.

    (If you had a movie camera set up, you could film the flight of the ball and deduce from this how long the ball was in the air, and then compare it with your calculation.)
     
  6. Sep 18, 2011 #5
    Re: Motion

    How about equation 5?

    The acceleration is local acceleration of gravity g. At this point one must remember that while these quantities appear to be scalars, the direction of displacement, speed and acceleration is important. They could in fact be considered as uni-directional vectors. Choosing s to measure up from the ground, the acceleration a must be in fact −g, since the force of gravity acts downwards and therefore also the acceleration on the ball due to it.

    At the highest point, the ball will be at rest: therefore v = 0. Using the fifth equation, we have:

    s= {v^2 - u^2}/{-2g}.

    Substituting and cancelling minus signs gives:

    s = {u^2}/{2g}.

    And therefore I can calculate the initial velocity u.

    Thanks in advance!
     
    Last edited: Sep 18, 2011
  7. Sep 18, 2011 #6

    NascentOxygen

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    Re: Motion

    (5) is useful only for the vertical component. You are throwing the ball horizontally, so (5) is of no use.
     
  8. Sep 18, 2011 #7
    Re: Motion

    You said that I know u which is the initial velocity, that is the component I'm looking for.
     
  9. Sep 18, 2011 #8

    NascentOxygen

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    Re: Motion

    You are trying to find u in the horizontal analysis. In the vertical analysis, u is known and is 0, because the ball is being thrown horizontally.
     
  10. Sep 18, 2011 #9
    Re: Motion

    Is a=g in equation 3? And how do I simultaneously determine t? Is it a simultaneous equation? Equation 4 which is the horizontal analysis does not contain u which is the initial velocity, how do I determine the initial velocity then?
     
    Last edited: Sep 18, 2011
  11. Sep 18, 2011 #10

    NascentOxygen

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    Re: Motion

    Yes, for the vertical motion.
    Solve 2 equations simultaneously; there being only one unknown there, t.
    The horizontal u is what you are trying to find. :smile:
     
  12. Sep 18, 2011 #11
    Re: Motion

    I really do not understand what you mean, the equations are:

    11htif7.jpg

    Equation three is the vertical on while equation four is the horizontal one,

    u= initial velocity
    v= final velocity

    If I'm throwing something horizontally, how do I determine the initial velocity. Equation four states only the final velocity and u=0 in equation three.
     
  13. Sep 18, 2011 #12

    NascentOxygen

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    Re: Motion

    No they're not. (3) and (4) are one and the same. For vertical motion, a=g. For the horizontal motion, a=0. The motion we are concerned with here involves both a horizontal component, and a vertical component. They need to be considered separately. Gravity affects only the vertical component. Nothing affects the horizontal component, so horizontal velocity remains unchanged (until the ball hits the dirt). You can use any equation anywhere, so long as you use the applicable value/s.

    That's the whole exercise here. We've made that your maximum speed of throwing the ball. That's the only unknown, provided you do measure the distance the ball goes before it hits the ground.

    Go back to my first post. That tells you how to go about finding u (horizontally).
     
  14. Sep 18, 2011 #13
    Re: Motion

    So how do I really find the initial velocity with regards to equation 3 and 4. I really do want to understand this problem and solve it. Sorry if I'm asking too much, I got a bit confused.
     
    Last edited: Sep 18, 2011
  15. Sep 18, 2011 #14
    Re: Motion

    I don't understand how I shall deal with the vertical analysis and the horizontal analysis and how to use equation (3) and (4) to determine the initial velocity. Please help!

    Thanks in advance!
     
  16. Sep 18, 2011 #15
    Re: Motion

    Are you there?
     
  17. Sep 18, 2011 #16
    Re: Motion

    11htif7.jpg

    Equation 3 which should be applied vertically contains u = initial velocity
    Equation 4 which should be applied horizontally contains v = final velocity

    How can I determine u which is the initial velocity in equation 4 that does not contain u (initial velocity).

    Thanks in advance!
     
  18. Sep 20, 2011 #17

    NascentOxygen

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    Re: Motion

    Oh, I hadn't distinguished the u from the v. Forget about (4), apply (3) separately for both horiz and vert components.

    horiz: you measure s, need to find u, know a=0, and leave t as t.

    vert: you measure s (vert distance from your hand to the ground as you release the ball), u=0, a=g, and leave t as t.

    Rearrange one of these as t=....
    and substitute that for t into the other equation. The only unknown then is u, that's the horizontal speed which you are wanting to determine, so you can rearrange the equation to find u.

    Are you doing practical trials of this on a playing field? If so, repeat half a dozen times, and see how well it works out. The key is to throw the ball precisely horizontally.
     
  19. Sep 20, 2011 #18

    gneill

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    Re: Motion

    The question is a bit vague. Do they mean the maximum velocity as might be produced by hitting it with a racket? By throwing it? Firing it out of a canon?

    Is that the entire wording of the original question?
     
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