Kinematic problem

  1. 1. The problem statement, all variables and given/known data
    An object is thrown up from the top of a building 50 m high. It rises to a maximum high of 20 m above the roof.

    A) when does it land on the ground
    B) At what velocity does it land ?
    C) When is it 20 m below the roof?


    2. Relevant equations

    The 4 kinematics equations the ones I used where

    Vf = Vi+ at

    vf ² = vi ² + 2ax



    3. The attempt at a solution

    Well I am completely lost now. I don't know if we can do this in one shot or two.

    I first found out how much time it takes for the ball to attein maximum height which gave me 2.02 seconds.

    After I calculate the time it takes to go from that high to the ground and in total I got 6.35 s

    but the textbook gave me 5.84

    as for the velocity I had - 42 m/s

    but the text book says -37 m/s :S


    I kept alot of decimals so I don't think the mistake is there
     
    Last edited: Sep 30, 2007
  2. jcsd
  3. So for the first path you should get that

    [tex] t_1 = \frac{\sqrt{2gy_{high}}}{g}[/tex]

    For the second path you should use the distance equation

    [tex] t_2 = \sqrt{\frac{2 y_{net}}{g}}[/tex]

    These two times will give you just about everything. Is this what you got?
     
  4. Step 1 : Decide the reference point for measuring the displacements i.e. y = 0 point and the +ve direction for displacement. It could be the ground in which case the initial displacement will be +50m and final displacement 0. Alternatively it could be the top of the building with upwards displacement +ve. In that case initial displacement will be 0 and final displacement -50m. Let us assume the second alternative.
    Remember : velocities and accelerations in the direction of +ve displacement are +ve and otherwise -ve.

    Step 2 : First determine u by using v^2 - u^2 = 2as. At the highest point v = 0. Displacement s = +20m and acceleration a = -g = 9.8 m/s2 (-ve because g is downwards).

    Step 3 : Use s = ut + 1/2 at^2 to determine time of hitting the ground. s = -50m and a = -g

    Step 4 : To determine the landing velocity use v^2 - u^2 = 2as. u as determined above, a = -g and s = -50m

    Step 5 : When is it 20 m below the roof : s = -20m. Which equation will you use?
     
  5. Good advice by Vijay, the only thing that I would add is to not put in any numbers until the absolute last moment possible. Do everything algebraically.
     
  6. Hmmm am not so familiar with those symbols is it,

    u = initial velocity

    Still looking at it for now
     
  7. Well thank you very much , this was pretty helpful I was never using a reference point correctly before, just got the last answer with Vf=Vi +at
     
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