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Projectile Motion pitching speed

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizontally from an elevation 4.00 m above the ground. The ball lands 30.0 m away.

    Part A:
    What is his pitching speed?

    Part B:
    As you think about it, you’re not sure he threw the ball exactly horizontally. As you watch him throw, the pitches seem to vary from 5° below horizontal to 5° above horizontal. What is the range of speeds with which the ball might have left his hand? Enter the minimum and the maximum speed of the ball.


    2. Relevant equations
    Sf=Si+Vis*T+.5aT^2


    3. The attempt at a solution

    For Part A I did the following:

    Yf=Yi+Viy*T+(.5)(g)T^2

    0m=4m+(0m/s)(T)+(.5)(-9.8m/s^2)(T^2)
    T=.9035

    Xf=Xi+Vix*T+(.5)aT^2

    30m=0m+Vix(.9035 seconds)+(.5)(0m/s^2)(.9035^2)
    Vix=33.204m/s


    Part B:

    I could only think to do the following...

    cos(5)=33.2/h

    h=33.3268....

    I don't really know what else I can do...Please help
     
  2. jcsd
  3. Feb 6, 2010 #2
    You need to remember the motion takes place in two dimensions. In the horizontal direction x = x_0 + v_xo*t. In the vertical direction
    y = y_0 + v_yo*t + .5*a*t^2.
     
  4. Feb 6, 2010 #3
    I dont follow. My part A is already correct.
     
  5. Feb 6, 2010 #4

    rl.bhat

    User Avatar
    Homework Helper

    Your part A is correct.
    For part B, the horizontal component of velocity will be
    vix= vo*cos(theta) and vertical component of viy will be vo*sin(theta)
    While calculating t, if the ball is projected up, take g -ve. And if the ball is projected down take g +ve.
     
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