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Projectile motion of a tennis ball

  1. Sep 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A tennis player hits a ball at ground level, giving it an initial velocity of 21.0 m/s at 58.0 deg above the horizontal. What is the ball's velocity at the highest point?

    2. Relevant equations
    Vx=Vocos(theta)
    Vy=Vosin(theta)
    y=1/2-g(t^2)+Vy(t)
    V-Vo/a = t



    3. The attempt at a solution

    I've completed all of the steps up to this point and arrived at the conclusion that the velocity of the ball at the highest point must be zero, theoretically speaking, of course. When I input 0 m/s, the answer was marked as incorrect. What am I missing. Final velocity must be zero since the balls velocity slows down to zero when it's at the maximum distance. It also asked for the acceleration at the highest point and theoretically speaking, the answer was 9.8 m/s^2, which is consistent with theory. What gives?!
     
  2. jcsd
  3. Sep 16, 2007 #2
    Hehe stumped arent you?
    But think.There are 2 velocities.Vx and Vy.Which one is zero?Both or one of those?
     
  4. Sep 16, 2007 #3
    Yes, I am completely stumped. If they're asking for velocity at highest point, naturally I'd think Vy. I didn't think Vx could even have a "highest" point?! Only Vy has a velocity of zero. As for Vx, what could what could that possibly be?

    Here's what I'm thinking now...

    Using equation v = Vo + at

    Substitute initial velocity of x (Vx) which is 11.1m/s, the acceleration of 9.8 m/s^2 and the time it takes to reach the balls max heigh which is 1.82s ...

    so, v = 11.1m/s + 9.8m/s^2(1.82s)
    v = 28.9 m/s

    Is that correct, or am I fooling myself? Doesn't make sense to me!:rolleyes:
     
  5. Sep 16, 2007 #4
    NO.You got the question wrong.They asked the V and not the Vy.At any point V is given as under root Vy^2 + Vx^2.

    And what is the equation u put above.Makes no sense.you cant combine Vx with Ay.Vx is always constant.Only Vy changes with time.
     
  6. Sep 16, 2007 #5
    Vo doesn't equal Vx
     
  7. Sep 16, 2007 #6
    Well, does Vo the originally stated 21.0m/s? That would mean the final velocity is 38.8 m/s... still not making sense. Please clarify, someone!

    Originally Posted by Idealism_Theory
    Yes, I am completely stumped. If they're asking for velocity at highest point, naturally I'd think Vy. I didn't think Vx could even have a "highest" point?! Only Vy has a velocity of zero. As for Vx, what could what could that possibly be?

    Here's what I'm thinking now...

    Using equation v = Vo + at

    Substitute initial velocity of x (Vx) which is 11.1m/s, the acceleration of 9.8 m/s^2 and the time it takes to reach the balls max heigh which is 1.82s ...

    so, v = 11.1m/s + 9.8m/s^2(1.82s)
    v = 28.9 m/s

    Is that correct, or am I fooling myself? Doesn't make sense to me
     
  8. Sep 16, 2007 #7

    Well, I tried that. It was incorrect! Now what do you suggest. I definitely didn't see that as the answer either.

    v = sqrt vx^2 + vy^2

    v = sqrt 11.1^2 + 17.8^2

    v = 30 m/s (which is marked as incorrect)
     
  9. Sep 16, 2007 #8
  10. Sep 17, 2007 #9
    Put Vy = 0 man. so that gives u 11.1 m/s as the answer.
     
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