1. Limited time only! Sign up for a free 30min personal 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!

Stone Throw Kinematics Question

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A stone is projected vertically upwards from the top of a tower. It is found that its speed at a distance of x meters below the point of projection is twice its speed at the same distance above that point. Show that the stone rises a distance 5x/3metres above the top of the tower.


    2. Relevant equations

    v2 = u2 + 2as

    3. The attempt at a solution

    Let the velocity from the projection point upwards be v1
    Let the velocity from the projection point downwards be v2

    v1 = u2 - 19.6x
    v2 = u2 + 19.6x

    Since V1 = 2v2

    u2 - 19.6x = 2u2 + 39.2x

    -58.8x = u2

    I am stumped.
     
  2. jcsd
  3. Jan 30, 2010 #2

    rl.bhat

    User Avatar
    Homework Helper

    When the stone reaches the maximum height h, its velocity is equal to zero.
    From that height if the stone is dropped, its velocity at a distance (h-x) is
    v^2 = 2*g*(h-x). ....(1)
    According to the problem, its velocity at a distance (h+x) is 2v and it is equal to
    (2v)^2 = 2*g*(h+x).....(2)
    Divide eq.2 by eq.1 and solve for h in terms of x.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stone Throw Kinematics Question
  1. Throwing A Stone (Replies: 4)

Loading...