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March 25, 2017

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A ball is thrown from the top edge of a building with initial velocity components of 15m/s(up) vertically and 20m/s horizontally. It strikes ground 140m from the base of the building. What is the height of the building?

I tried using V2^2 = V1^2 + 2ad and it gave me 11.25 which I know is wrong. How would I calculate this?

  • Physics - ,

    The ball lands 140m from the base of the building. The balls initial horizontal velocity is 20m/s.
    140m = (20m/s) * t
    Solve for t.
    Since the time the ball is in flight is now known.
    Plug that into the "complete" formula for distance:
    d = (1/2)*(g)*(t^2) + v*t + h
    Let distance 0 be ground level.
    initial vertical velocity is v=15m/s
    h is the initial distance above the ground which is the height of the building.
    0 = (1/2)*(g)*(t^2) + v*t + h
    Solve for h.
    Remember that the initial velocity is up and the acceleration of gravity is down (watch the signs)

  • Physics - ,

    I helped you with this question yesterday and gave the exact same info as Quidditch Why didn't you use the info I gave?

  • Physics - ,

    Thanks! I saw it and used it but I'm not sure if I did it right. I got 140 for the height of the building and I don't think that's right.

  • Physics - ,

    I got something in that range, but I believe your answer has more error than I would expect.
    What did you get for time that the ball was in the air?
    What numbers did you plug into your final equation?

  • Physics - ,

    For time t = d/v
    140/20
    t= 7

    0 = 1/2(-10)(7^2)+(15)(7)+h
    -h = -140
    h = 140

    We were asked to use 10 for our acceleration.

  • Physics - ,

    You got it!
    Using 10m/(s^2) for gravity, your answer is correct.

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