1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinematic problem

  1. Sep 30, 2007 #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. Sep 30, 2007 #2
    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. Sep 30, 2007 #3
    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. Sep 30, 2007 #4
    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. Sep 30, 2007 #5
    Hmmm am not so familiar with those symbols is it,

    u = initial velocity

    Still looking at it for now
     
  7. Sep 30, 2007 #6
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Kinematic problem
  1. Kinematics problem (Replies: 5)

  2. Kinematics Problems (Replies: 4)

  3. Kinematic problem (Replies: 1)

Loading...