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Projectile Motion Question Please Help

  1. Jan 31, 2010 #1
    1. During a tennis match, a player serves the ball at 29.4 m/s, with the center of the ball leaving the racquet horizontally 2.41 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.

    2. I know that x = x_0 + v_0t + 1/2at^2(this can be applied in the y direction as well), y=(v_0sin(theta))t - (gt^2)/2

    3. I've looked at the problem for a good thirty minutes and I've gotten no where. Please help.
     
  2. jcsd
  3. Jan 31, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi MFlood7356! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Yes, use those equations, and remember that a = 0 for the x-direction …

    you know x, so find t, then find y.

    What do you get? :smile:
     
  4. Jan 31, 2010 #3
    Ahh okay I was confused on what the acceleration was because I was having a hard time picturing the problem. I ended up getting 0.408s for time. Now when solving for y I use -9.8 as the acceleration correct?
     
  5. Jan 31, 2010 #4
    Okay I used y = y_0 + v_0t + 1/2at2 to get 1.59m then 1.59m - 0.900m to get 0.694. Is that right?
     
  6. Jan 31, 2010 #5
    I put in 0.694m as my answer and that is correct. I attempted to go through and do part b but when I entered the answer it was incorrect. Here's what I did.

    Vy= 29.4m/s(sin5) = 2.562 Vx = 29.4m/s(cos5) = 29.288
    t = 12/29.288 = 0.410s
    y = 2.41m + 2.562(0.410) - 1/2(9.8)(0.410)2 = 2.63673 - .900 = 1.74m
     
  7. Jan 31, 2010 #6

    tiny-tim

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    "below the horizontal" :wink:
     
  8. Jan 31, 2010 #7
    Would that be negative 5 degrees then?
     
  9. Jan 31, 2010 #8

    tiny-tim

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    Of course. :smile:
     
  10. Jan 31, 2010 #9
    Okay I got the correct answer thank you!
     
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