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Projectile motion of a baseball problem

  1. Feb 23, 2008 #1
    1. The problem statement, all variables and given/known data
    a baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23.0m/s. If the ball was caught 42 m from the thrower, how long was it in the air? how high did the ball travel before being caught?

    2. Relevant equations


    3. The attempt at a solution

    using x=(VicoxA)t --> 42/(23*cos25) i got 4.32
    and than using y=(23*sin25)4.32+1/2(-9.81)(4.32)^2 i got -49.5 >>y=(VisinA)t+1/2a(t)^2)
    but in the back of the book it says its wrong so can someone please tell me what i did wrong?
     
  2. jcsd
  3. Feb 23, 2008 #2
    using x=(VicoxA)t --> 42/(23*cos25) i got 4.32
    and than using y=(23*sin25)4.32+1/2(-9.81)(4.32)^2 i got -49.5 >>y=(VisinA)t+1/2a(t)^2)
    but in the back of the book it says its wrong so can someone please tell me what i did wrong?[/QUOTE

    The formula for projectile motion is the Y direction is going to be delta y=Vt -1/2gt^2 assuming gravity is -9.8. As a result of this double negative you end up adding the latter half of the equation where it appears you have subracted by adding a negative. Give that a try and see if it gives you a better answer
     
  4. Feb 23, 2008 #3
    well....look at your second step: you are not calculating the hightest point...(and i'm not sure but -49.5 is not right: you are saiyng that y=-49.5?? the ball must be underground...xD)

    hint: In the hightest point, the ball as no speed in y axis...^_^
     
  5. Feb 23, 2008 #4
    ooo thanks a lot!
     
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