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Homework Help: 2D Motion Revision

  1. Feb 4, 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 3.00 m above the ground. The ball lands 30.0 m away.

    A) What is his pitching speed?

    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 degrees above horizontal. What is the range of speeds with which the ball might have left his hand?

    2. Relevant equations

    d = vi*t + (1/2)a*t2

    3. The attempt at a solution

    Problem A was completed without issue. But I'm having trouble working the possible angles into the answer.

    I know that vx = vcos(5) and vy = +/- vsin(5)

    I tried to isolate time, given that the horizontal distance was traveled in the same time it took the ball to hit the ground (obviously...).

    vx = vcos(5) = .99v
    vy = -vsin(5) = .08v

    dx = vcos(5)t
    30 = vcos(5)t
    t = 30/(vcos(5))

    3 = -vsin(5) - (1/2)(9.8)(t^2)
    t = sqrt((-vsin(5) - 3)/4.9)

    Now I would usually set the equations equal to each other at this point, but the result system seems too messy to be done by hand, so I feel like I'm missing a much easier solution.

    Is my only option to try to find the intercept for the equations? And if the ball was thrown upwards, my equations would be the same except for the negative in from of the sin portion, correct?
  2. jcsd
  3. Feb 4, 2010 #2


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    The traditional procedure for dealing with this sort of thing is to solve the x-equation for t and replace t in the y-equation. The algebra is much more tractable.
    Last edited: Feb 4, 2010
  4. Feb 4, 2010 #3
    That doesn't exactly make the math any easier.

    3 = -vsin(5) - (1/2)(9.8)(302/v2cos2(5))

    That doesn't exactly make the isolation of v any easier...
  5. Feb 4, 2010 #4


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    The equation that you have for v doesn't look right. The y equations should be

    0 = 3 + v*t*sin5o-0.5*9.8*t2

    0 = 3 - v*t*sin5o-0.5*9.8*t2

    Either of these equations says that at t=0 the ball is at y=3 m.
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