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Speed of stone thrown

  1. Aug 29, 2003 #1
    A stone is thrown vertically upward. On its way up it passes point A with speed v , and point B, 3.00 m higher than A, with speed 1/2 v . Calculate the speed v
    Here is how I tackled this problem. Let v = final velocity, and u = initial velocity.

    v^2 = u^2 - 2gd eq.1

    v^2/4 = u^2 - 2g(d+3) eq.2

    Rearrange eq. 2 to obtain
    v^2 = 4u^2 - 8gd - 24

    Solve u
    4u^2 - 8gd - 24 = u^2 - 2gd

    3u^2 = 6gd + 24

    u^2 = 2gd + 8

    v^2 = 2gd + 8 - 2gd
    = 8
    v = 2 [squ] 2 ms^-1

    But the book says the answer is 8.85 ms^-1. So what am I doing wrong?
     
  2. jcsd
  3. Aug 29, 2003 #2

    Chi Meson

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    Homework Helper

    There' only one equation.

    When the stone passes point A, that's position "zero" with initial velocity "v". At point B, the displacement is now 3 m and final velocity is "v/2." You chose the right equation. Along the way to the answer you shoul get "(3v^2)/4 = 6g "
     
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