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Maximum height reached by a projectile

  1. May 15, 2012 #1
    http://dl.dropbox.com/u/33103477/gravity.png [Broken]

    I have worked out the first two bits. (So you can assume that I have that) But I can't figure out how to work out the maximum height reached.

    I know there I can equal the KE and the PE to work out max height, but that doesn't look like it'll work here.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 15, 2012 #2

    vela

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    Integrate v(t) to find y(t).
     
  4. May 15, 2012 #3

    Ray Vickson

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    The KE/PE argument will not work because of the presence of air drag. However, you don't need it. You have [itex] v = dy/dt.[/itex] Do you recall the conditions for a maximum of y(t)?

    RGV
     
    Last edited by a moderator: May 6, 2017
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