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Projectile Motion with only the vertical/ horizontal height and angle given

  1. Nov 3, 2011 #1
    I need help! I understand regular projectile, but when it comes to using an angle it drives me crazy!
    The vertical distance is 15.01 m
    Horizontal distance is 21.84 m
    Angle is 45°
    I figured the time to be 2.1 sec
    I need the velocity
    And VsubF is 0...right?
    I just need the help guys, thanks
     
  2. jcsd
  3. Nov 3, 2011 #2
    Which quantities are given and which did you find out (and how)?
    It would be nice to show the problem you are trying to solve.
     
  4. Nov 3, 2011 #3
    Don't worry about the angle. V and H motion are still independent. Break the trajectory in half, and let's check if your time is correct. A falling object falls as d=0.5gt^2, this one fell 15m, if G≈10m/s^2 then this one fell:
    15m=0.5*9.8(m/s^2)*t^2 --> t=√15/4.9=1.7496s
    Vhorizontal=Δx/t=21,84m/(2*1.7496s)=6.241m/s (the time is doubled because the time in the first step is only for the falling part of the motion.
    Vinitialvertical=Vhorizontal
    Vinitialtotal=Vhorizontal*√2 by Pythagoras.

    If those aren't what you're looking for, you need to describe the problem more.
     
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