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Time vs Height - EASY but easily forgotten :/

  1. Jan 21, 2009 #1
    1.The problem statement, all variables and given/known data
    One day, while working on a downtown construction site, Eugenia accidentally dropped her cell phone into the large excavation on the site. Fortunately, the cell phone landed on a pile of soft earth and was not damaged. She asked Tony, the construction manager, who was working nearby to throw the cell phone back up to her.

    The height of an object thrown vertically up is modeled by

    s(t) = s0 + v0t - 1/2gt^2

    where s0 is the initial height above the ground, v0 is the initial velocity, g is the acceleration due to gravity, t is the time, and s(t) is the height above the ground at time t.

    If Tony is 10m below ground level, and the acceleration due to gravity is 9.8m/s^2, with what initial velocity must he throw the cell phone for it to reach Eugenia? Explain any assumptions you make and show your calculations, graphs, or both to justify your answer.

    2. Relevant equations

    h(t) = t^2 + t + c

    s(t) = s0 + v0t - 1/2gt^2

    3. The attempt at a solution

    Last edited: Jan 21, 2009
  2. jcsd
  3. Jan 26, 2009 #2

    Gib Z

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    Homework Helper

    Hello Synix, welcome to PF!

    Are you familiar with the equation [tex] v_f^2 = v_i^2 + 2a\Delta s[/tex] ?

    Thats the equation we need for this one. After Tony throws up the cellphone, when it reaches Eugenia it should be at the top of its flight. At this point, was is its velocity? I'm sure you can figure it out from here.
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