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Math Help - Stone throwing velocity problem

  1. #1
    Member mybrohshi5's Avatar
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    Unhappy Stone throwing velocity problem

    A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 12 feet per second. A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? B. At what time does the stone hit the ground? C. What is the velocity of the stone when it hits the ground? I have tried attempting this problem multiple ways and i still cant get the correct answers Can anyone help me. Thank you.
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  2. #2
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    Well, what you know is that a= -32. Using this you can find the velocity at time t by integrating and solving for the integrating constant. And from that you can find the position at time t doing the same thing you did for velocity. You solve for the integrating constants using your initial conditions.
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  3. #3
    Member mybrohshi5's Avatar
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    What equation would i use for this though?
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  4. #4
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    Quote Originally Posted by mybrohshi5 View Post
    A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 12 feet per second. A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? B. At what time does the stone hit the ground? C. What is the velocity of the stone when it hits the ground? I have tried attempting this problem multiple ways and i still cant get the correct answers Can anyone help me. Thank you.
    you should have at least been exposed to the vertical motion equation for position and how it is derived ...

    h(t) = h_0 + v_0 t - 16t^2

    you were given v_0 = 12 ft/s and h_0 = 925 ft (a roof height ?)

    part (a)

    evaluate h(2)

    part (b)

    set h(t) = 0 and solve for t

    part (c)

    h'(t) = v(t)
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