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A ball is thrown from the roof of a building

  1. Dec 6, 2005 #1
    It is s(t)=245+24.5t-4.9t(sqrd) meters from the ground (t) after it is thrown. s(t) = displacement function that describes the position of the ball from the ground at time t.

    Determine height of building, time it takes the ball to reach the ground and the velocity when it hits the ground.

    I've been messing around with first, second derivatives and breaking down the original equation into t=10 and t=-5 but I don't know where to go from here.

    Help apperciated.

    i should mention i took physics 2 years ago and dont remember much of it
    Last edited: Dec 6, 2005
  2. jcsd
  3. Dec 6, 2005 #2


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    Well since the idea of negative time here is worthless, how could you use that t=10?
  4. Dec 6, 2005 #3
    Well I plug it in the equation and get 980???
  5. Dec 6, 2005 #4


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    What equation did you put t=10 into?
  6. Dec 6, 2005 #5
  7. Dec 6, 2005 #6


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    Recalculate that. What do you think the position will be after 10 seconds?

    You wouldn't think that after 10 seconds, the object is 980 meters off the ground...
  8. Dec 6, 2005 #7
    Of course I woudn't that's why it doesn't make any sense to me.
    I think im fundementally not getting some kind of simple concept here.
    1st derivative when 10 is plugged in i get 1.47
  9. Dec 6, 2005 #8


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    Well you were doing the calculation wrong.

    245 + 24.5(10)-(4.9)(10)(10)=0 which is the position at t=10.

    What equation do you use for finding the velocity in the y direction?
  10. Dec 6, 2005 #9
    v(t)=s dirvative(t)

    I don't know. I'm not taking physics. I've covered my calculus books chapter and coudnt find anything.

    I'm gonna rest on it, thanks for you help though.
  11. Dec 6, 2005 #10


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    Where is the ball at t=0?

    At what time does s(t) = 0 and where is the ball then?

    How do you find the function which discribes the velocity as a function of time?
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