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Homework Help: Physics Problem using Equations of Motion

  1. May 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A brick is dropped from a hot air ballon which is ascending at 5ms^-1. The height of the balloon above the ground at the time of release of the brick was 30m.

    a. Determine the time that it took the brick to hit the ground.
    b. At what height was the balloon above the ground at the instant the brick hit the ground?
    c. What was the velocity of the brick when it hit the ground?
    d. Sketch a graph of the velocity against time for the brick over this journey.

    2. Relevant equations

    1. v=u+at
    2. v^2=u^2+2as
    3. s=ut+1/2at^2
    4. v=s/t

    3. The attempt at a solution

    For this question I specified up direction as negative and down direction as positive.

    For (a)

    I decided to use the formula, s=ut+1/2at^2, where I substituted s for 30, u for -5 and a for 9.8. When rearranged this gave me 4.9t^2-5t-30=0. Using the quadratic formula I then got the answer t=-2.53 and 2.53.

    I would have thought that these values would have been different however as after being released from the balloon, the brick would surely have risen slightly before gravity accelerates it towards ground meaning the positive value should be greater than the negative value.

    For (b)

    I used the s=ut+1/2at^2 formula with s=30, u=-5 and a=0 values of the ballon to prove that after six seconds the balloon was 30m above the ground. I got a t=-6 answer for this though and thus maybe I did something wrong, it should be positive right? The I used the s=ut+1/2at^2 formula using -8.53 (-6-2.53)*-5 to give me the displacement 42.65m.

    For (c)

    I used v=u+at with the values, u=-5 a=9.8 and t=-2.53 to give me the velocity 19.794 ms^-1 down.

    I will attempt (d) after finishing a,b and c.
  2. jcsd
  3. May 19, 2013 #2


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    These results are wrong. How do you solve a quadratic equation?

    The formula for the displacement you used is correct, and it accounts for the small intial rise of the brick. Try to find the displacement at t=0.1 s.

    The balloon was at 30 m height when the brick was dropped out. You can consider that instant as zero of the time. The balloon ascends with 5/s speed so it rises by how many m-s during the time the brick falls? What will be the final height?
    But you need the correct time of fall first.

  4. May 19, 2013 #3
    I believe I solved for time wrong.

    There should be a way to solve for t using the equations I listed, without using the quadratic formula.

    Why solve for t=0.1? It equals -30.451.
  5. May 19, 2013 #4


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    You have to solve the quadratic equation to get the time. Use the quadratic formula or completing the square - you certainly have learnt some of them?

    To find the height at t=0.1 just illustrates that the formula s = ut+a/2t2 works both for positive and negative displacements.

  6. May 19, 2013 #5
    I have decided to change it to make up positive and down negative now. Makes more sense to me now.

    For (a) I just realised this could also have been done by using v^2=u^2+2as to find v=24.74ms^-1 and then using v=u+at, to find t, being -2.02

    For (b) Using s=ut+1/2at^2 with s=5(6+2.02) I get s=40.1m for the height of the balloon the instant the brick hit the ground.

    For (c) The answer is the v=24.76ms^-1 I previously found.

    Now how would I set out the axis for (d)??
    Last edited: May 19, 2013
  7. May 19, 2013 #6


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    You are right you can find the time without the quadratic formula, but your result is not correct. Is it possible that the brick reaches the ground in negative time? Before it has been dropped out from the balloon?
    Velocity has sign, positive when up and negative when moving down. Is that v=24.74 m/s upward or downward? Positive or negative?

  8. May 19, 2013 #7
    It is 24.76ms^-1 downwards.
  9. May 19, 2013 #8


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    What values you have to plug in for u and v into the equation v=u+at?

  10. May 19, 2013 #9
    If you rearrange v=u+at, you get t=(v-u)/t

    I substituted 24.76 for v, and 5 for u.

    Just while typing this I realise that I should be substituting -24.76 and thus getting t=3.03.

    This will thus change my answers for (a) and (b). Thanks for calling my attention to that.
  11. May 19, 2013 #10


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    Choose t (time) for the horizontal axis and v (velocity) as vertical.

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