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Finding time and range for a projectile launched from a cliff

  1. Jun 14, 2007 #1

    exi

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    Traced this down to an error in my formula as I'd copied it down; it's (sqrt(Voy^2 + 2gh)+Voy) / g, not (sqrt(Voy^2 + 2gh)-Voy) / g. Thanks though!

    1. The problem statement, all variables and given/known data

    A 0.57 kg projectile is fired into the air from a cliff that's 13.9 m above a valley.
    Initial velocity = 7.97 m/s
    Angle: 51° above horizontal.
    Acceleration = g

    1: How long is the projectile in the air?
    2: How far from the bottom of the cliff does the projectile land?

    2. Relevant equations

    I tried to use (sqrt(Voy^2 + 2gh)-Voy) / g to find time after breaking down the initial velocity into axial components, but this yields 1.1669 s - obviously incorrect (and verified as such).

    3. The attempt at a solution

    My shot at the first portion of this is above, since the equation was used in lecture as a way to find the time (in order to find the range) of an object fired at an angle from an elevated position. Just over one second doesn't check out logically, but I'm not sure why a given formula isn't yielding proper numbers.
     
    Last edited: Jun 14, 2007
  2. jcsd
  3. Jun 14, 2007 #2

    Astronuc

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    Staff: Mentor

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