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Formula for impact velocity at an angle

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched from a cliff that is 84m tall with an initial speed of 32m/s at an angle of 25 degrees above the horizontal. The object is in the air for a total of 5.74s. What is its impact velocity and at what angle?


    2. Relevant equations
    Vfy = Voy + ayt

    Vf^2 = Vfx^2 + Vfy ^2

    tan(theta) = (Vfy)/(Vfx)


    3. The attempt at a solution
    I have that the answer is 51.7m/s at an angle of 55.9 degrees below the horizontal. However, when I attempt to solve, my answers don't match up:

    Vfy = Voy + ayt
    = 0 + (-9.8m/s^2)(5.74s)
    = -56.25

    Vf^2 = Vfx^2 + Vfy^2
    = (32)^2 + (-56.25)^2
    = 1024 + 3164.06
    = 4188.29
    Vf = 64.71m/s <------ incorrect

    Then when I plug in to find the angle of impact:

    tan(theta) = (Vfy)/(Vfx)
    = (-56.25)/(32)
    = -1.7578
    tan^-1 (-1.5758) = -60.36 degrees below the horizontal <------- also incorrect.

    I'm not sure what I'm doing wrong. Any suggestions??

    I also need to calculate the peak height of the projectile; I first used:
    t = (Voy)/g
    = 32sin25/(-9.8m/s^2)
    = 13.5/(-9.8)
    = 1.37s

    then y = (1/2)at^2 + Voyt
    = (-4.9m/s^2)(1.37s)^2 + 13.5m/s(1.37s)
    = -9.2 + 18.5
    = 9.3m <-------- this answer cannot be correct, not sure what I'm doing wrong
     
    Last edited: Oct 9, 2009
  2. jcsd
  3. Oct 10, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi laurmaso! Welcome to PF! :wink:
    Sorry, but this makes no sense :redface:

    Voy isn't 0, and you haven't used the 84m.

    Try again. :smile:
     
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