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Thrown Upwards with one variable!

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    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 (a) the speed v and (b) the maximum height reached by the stone above point B.

    2. Relevant equations
    Time in the sky= Initial Velocity/9.8m/s^2
    Maximum Height= xmax = Initial Velocity^2/ 2 x (9.8/m/s^2)

    3. The attempt at a solution

    I've been experimenting and can't seem to find an answer. I've tried using x+3+y to fill in for the maximum height, but I'll always end up more than one variable to solve for. Because deceleration would be at gravity, 9.8m/s^2, I was trying to figure out a way to even calculate a time, but without an initial velocity, it seems hopeless.

    Thanks for the help :)
  2. jcsd
  3. Sep 15, 2007 #2


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    Homework Helper

    first check your relevant equations... they don't look right to me although they are written in some weird notation. check out the kinematics equations for constant acceleration at say
    should be a good start (although no one should really take wiki seriously :smile:)
  4. Sep 15, 2007 #3


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    Homework Helper

    oh.. and of course, welcome to PF.
  5. Sep 15, 2007 #4
    Thanks for the warm welcome, and I'll be sure to look it over :)
  6. Sep 16, 2007 #5
    In all such problems be careful of the sign (+ or -) of different quantities. First decide the +ve direction for displacement. If the object is thrown upwards, it is logical to have upward displacement as +ve. However, if the object is thrown upwards from top of a tower, there is an option to have upward or downward displacement as +ve. Then, the velocities and accelerations in the +ve direction are also +ve. Acceleration due to gravity is always downwards. Hence, it will be +ve or -ve depending upon the direction chosen as positive. In the present case it will be -ve as upwards is +ve.

    Consider motion of the stone from point A to point B. Use equation V^2 - Vo^2 = 2as and substitute Vo = V, V = V/2, s = 3m and a = - 9.8 m/s^2 and solve for V.

    To get the max height reached above B, again use the above equation with substitution : V = 0, Vo = V/2 (for which now we have a value) and a = -9.8 m/s^2 and solve for s.
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