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Homework Help: Projectile Motion of Ball Problem

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data

    A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.20 m/s at an angle of 19.0° below the horizontal. It strikes the ground 4.00 s later.
    (a) How far horizontally from the base of the building does the ball strike the ground?
    (b) Find the height from which the ball was thrown.
    (c) How long does it take the ball to reach a point 10.0 m below the level of launching?

    2. Relevant equations

    xf=xi+1/2(vxi +vxf)*t
    vxf2= vxi2+2*a(xf-xi)
    vf = vi + a*t
    vxi= vi*cos([tex]\Theta[/tex])
    vyi= vi*sin([tex]\Theta[/tex])
    h= vi2s*in2 ([tex]\Theta[/tex])/(2g), g= gravity constant 9.8 m/s2, h= max height
    R= vi2*sin(2[tex]\Theta[/tex])/g, R=max horizontal range

    3. The attempt at a solution
    I know how to do part (a). This is what I did
    vxi= vi*cos[tex]\Theta[/tex]
    vxi= 9.2*cos(19)= 8.70 m/s
    xf= 0+8.7*4 +0= 34.8 m

    But for part (b) and (c), im stuck, and I cannot do part (c) w/o doing part (b). Heres my attempt.
    What I did bascially is finding the time vyf = 0, which indicates the time of the maximum height. Then I used the yf equation to find the height of ball at that time. Then I just got stuck. Please help asap!
    Last edited: Sep 17, 2008
  2. jcsd
  3. Sep 17, 2008 #2


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    Homework Helper

    Think about the height equation Y = 1/2*a*t2

    But you also have an initial velocity down of 9.2*Sinθ added to the 1/2(9.8)*(4)2

    That yields your Total height Y = 9.2*Sinθ + 1/2(9.8)*(4)2

    Now use the same equation but rather than plug in T=4 and solve for Y, plug in Y=10 and solve for T.
  4. Sep 17, 2008 #3
    But I dont think that makes sense because you want the height of where you threw it from which was the upper story of the building, so u would have to find the time at which t equals same height as where you threw it from to find the initial height, but i have no clue on how to do that. Can you explain your process for part b again. Thanks
  5. Sep 17, 2008 #4


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    If you dropped it from that height of the building starting with 0 velocity, calling your height 0, and it takes 4 seconds to hit the ground how high is that?

    You will find that this distance would be given by Y = 1/2*g*t2

    Now if you threw it downward with an initial velocity will that make the distance longer if it still takes 4 seconds? Yes it will, by the value of the initial velocity times the time until you hit the ground. Hence that vertical velocity contribution to distance is the 9.2*Sine(19)*T where t=4 for this case.

    Edit: (I see that I omitted the *T to the velocity term the first time. )
    Last edited: Sep 17, 2008
  6. Sep 17, 2008 #5
    Okay now i get it, why didnt I think of it. lol. I have a question though. Why dont we put -9.8 m/s^2 for the gravity constant instead of a positive 9.8, because the ball is falling down?
  7. Sep 17, 2008 #6


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    So long as you are consistent with your signs, matching velocities and accelerations with what you call positive position, you can do it as you wish.
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