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Free fall physics question

  1. Sep 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A certain object is falling in free fall down a building. The object falls the last 30 meters in 1.5 seconds. What was the total height it was dropped at.

    2. Relevant equations

    s=(1/2)(g)t^2

    3. The attempt at a solution

    I've tried setting the total height to x+30 and solving for x.
     
  2. jcsd
  3. Sep 8, 2012 #2

    Actually you can use (1/2)gt² only if the initial velocity is 0. In the last 30m the initial velocity is not 0.

    You have 30 = V.t + gt²/2 where V is the velocity in h=30m.

    You find V=12.5
    So T = 12.5/10 + 1.5 = 2.75s
    Now you use gt²/2 and you get 37.81m
     
  4. Sep 8, 2012 #3
    lemme try

    suppose total height is h+30

    now let time be t(to cover h distance)
    h=1/2 gt2
    h+30=1/2g(t+1.5)2(it takes 1.5 sec more)

    solve for h and t u should find both
     
  5. Sep 9, 2012 #4

    CAF123

    User Avatar
    Gold Member

    It is best to split this problem into two separate parts. Consider the motion between the top of the building until the ball is at a height of 30m above the ground (first part). Then consider the motion from this 30m height to the ground (second part).

    If the total height fallen is [itex] H [/itex] then 30m above the ground, it has travelled [itex] (H -30)m. [/itex] It was dropped with zero initial velocity so for the first part of the motion, we have [tex] -(H-30) = -\frac{1}{2}gt^2,\,\, (1) [/tex] keeping consistent with signs.
    Now consider the second part. We want to determine the velocity at a height 30m above the ground. You know the time taken to fall this distance so using one of the kinematic relations you should be able to find this velocity.

    Now we have all the information required to find the time taken to travel [itex] (H-30)m [/itex]. This is found using another (different) kinematic relation. Sub this time into eqn (1) and you have your [itex] H. [/itex]
     
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