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!

Homework Help: Initial velocity when given distance and acceleration

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Material ejected to a height of 2.00 x 10^5m. Acceleration of gravity is 1.80 m/s^2, find initial velocity

    2. Relevant equations

    v (zero) = v - at.

    3. The attempt at a solution

    Seems as though I am missing something since I don't have time or velocity, yet I have distance. Guidance on steps would be appreciated.
  2. jcsd
  3. Sep 24, 2011 #2
    If you use like

    (Yf - Yi) = Vfx(t) - 1/2(Ay)t^2

    You know that when the material reaches it's max height (given in your problem), the velocity of it is zero right?
  4. Sep 24, 2011 #3
    Yes, makes sense at zero. Though the question is based on a volcano, so I am thinking initial velocity would be when it leaves the cone of the volcano.

    Unfortunatley, I am not familiar (or sure) how to use the formula you provided.
  5. Sep 24, 2011 #4
    It's this equation:

    [tex]\Delta y = V_{fy}t - \frac{1}{2}A_{y}t^{2}[/tex]

    Delta y is essentially y final minus y initial. If you let the tip of the volcano be 0 on the y axis, you would have the max height = final velocity times time minus one half acceleration due to gravity (given) times time squared. When the final velocity is zero then, you can solve for time to figure out the time when the material is at it's max.

    Then I think you can use the fact that:

    [tex]V_{fy} = V_{iy} + A_{y}t [/tex] using 0 for final velocity and the solution from the first equation for time to solve for initial velocity.

    Unless I am mistaken anyway. I am just learning this material myself.

    That first equation is just a derivation of the more common:

    [tex]\Delta y = V_{iy}t + \frac{1}{2}A_{y}t^{2}[/tex]
  6. Sep 24, 2011 #5
    BTW, great handle.

    First, how in the world are you able to enter these equations so clearly?

    If I understand, the first equation is now 200,000 = (Zero x t) - 1/2 (1.80 x t^2). Concerned unit of measure for A is not correctly represented in my step of the equation
  7. Sep 24, 2011 #6
    That's right. (Assuming I am correct to begin with). Solving for t will give you the time when the velocity is 0 (the time at the peak), then use the other equation with final velocity as 0 and time as the solution for t, and you should be able to solve for initial velocity.

    You can enter equations like this using LaTeX. Right click my equations and select show source to see how it's done. There is also a little latex helper in the text editor on this forum. Just click the capital sigma (E looking thing) while editing a post in advanced mode.

    edit: You have to give the code tags like [.tex] code [./tex] (without the periods) The latex helper thing gives itex tags, which are basically the same thing but they print in line with your text.
    Last edited: Sep 25, 2011
  8. Sep 25, 2011 #7
    Now I have a major appreciation when posters such as yourself take the time to present the equations in a clear format. Much more difficult than I imagined.

    VERY rough on my Algebra - the first equation would now be

    1.8 x 2t^2 = 200,000
  9. Sep 25, 2011 #8
    The first equation should be:

    [tex]\Delta y = V_{fy}t - \frac{1}{2}A_{y}t^{2}[/tex]

    So plugging in the given values you get:

    [tex]200000 = 0t - \frac{1}{2}1.8t^{2}[/tex]

    That's simply a quadratic equation right? You might get two answers for t, but remember there is no time travel.
  10. Sep 25, 2011 #9
    Quarkcharmer, you have been great. I have been helping my son, this is an online exam.

    He just told me that he was able to solve with your help and he got it correct.

    Really appreciate you taking the time to help out a couple of beginners, your patience is appreciated.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook