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A projectile equation, and then some

  1. Nov 15, 2007 #1
    1. Two 40g (.04kg) projectiles are launched horizontally at a point 2m above the ground. One lands 200m away and the other lands 250m. Find the PE and the KE of each projectile at launch.



    2. dx= vx
    dy=Vy(t)+1/2gt^2(t)
    KE= 1/2mv^2
    PE=mgh




    3.I tried plugging the variables into a drop equation, and came up with vi=0, no final velocity, d=200, t=6.4, and a= -9.8
    The variable I solved for was time. I'd like to know if I'm on the correct track, or if I'm doing this horribly, horribly wrong.
     
  2. jcsd
  3. Nov 15, 2007 #2
    Solve for the time it takes a projectile to drop 2m.

    Now, you should know a distance they traveled horizontally in a certain amount of time...

    You should be able to take it from there. Ill check back again later.
     
  4. Nov 15, 2007 #3
    If I use 2m as my distance, my time equals .64 seconds, when plugged into the equation d=(vi)(t)+1/2at^2

    The next step I did was to plug time into the equation Vx=dx/t
    For projectile 1, the vx is 312.5, and for projectile 2, the vx is 390.63

    I then found the KE for each projectile, and they were 6.25J and 7.81J, respectively.

    Potential energy should be the same for each projectile, correct?
    If PE=(.04)(-9.8)(2), then the PE for each projectile should be -.784

    End problem.
     
  5. Nov 19, 2007 #4
    I have another question:

    You are launching a large 200kg rock at the house of your Physics teacher. Your catapult applies 40,000N of force over the course of its 3m swing. If the stone releases so that it is heading at an angle of 30 degrees above the horizontal, then what is the minimum distance that her house would have to be away in order to be safe from the cannon?

    I started off with a drop equation to get time, which I calculated to be .78
    Then I calculated the dx to be 23.4

    But I'm a little lost from there.
     
    Last edited: Nov 19, 2007
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