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Projectile motion with limited variables to use, halp

  1. Mar 6, 2010 #1
    1. The problem statement, all variables and given/known data
    The starting hight is .51 metres above the ground, and the range is 4.651 metres. the angle of launch is 22 degrees above the horazontal. i need to find the initial velocity, and this was all i have to work with.
     
  2. jcsd
  3. Mar 6, 2010 #2
    i think it might be impossible to solve....i think you would at least need speed in that equation (initial velocity)
     
  4. Mar 6, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi xwater ! Welcome to PF! :wink:

    Start by calling the speed v, and writing (separate) equations for the x and y components of the motion.

    What do you get? :smile:
     
  5. Mar 7, 2010 #4
    hey, thanks for the replies, but i've tried making the x and y seperate, and i cant solve this. i end up with only a displacement for the x, and accelleration (gravity) and a displacement for the y. with any one more variable i'm sure i could solve this..errgg :( lol
     
  6. Mar 7, 2010 #5
    Ok, start by breaking the initial velocity into two components.

    In the x direction you have vcos22. In the y direction you have vsin22.

    Now start using kinematics.

    The x direction is simple. For a total flight time t,

    t = 4.651/vcos22

    The y direction is slightly more complicated. The projectile's path goes up, then back down to where it started and then continues for another 0.51 meters. This means its final displacement is 0.51 meters below its initial position. So,

    -0.51 = vsin22t - 0.5gt^2

    But we know t from before.

    -0.51 = vsin22*4.651/vcos22 - 0.5g(4.651/vcos22)^2

    -0.51 = tan22*4.651 - 0.5g(4.651/vcos22)^2

    -0.51 - tan22*4.651 = - 0.5g(4.651/vcos22)^2

    [-0.51 - tan22*4.651]/(-0.5g) = (4.651/vcos22)^2

    And I think you can see where it goes from there.
     
  7. Mar 7, 2010 #6
    The explaination by Atticus is nice and correct! :) I second it! :)
     
  8. Mar 7, 2010 #7
    Thanks a ton guys! Atticus, you blew my mind..lol. Now i just gotta go do it myself :)
     
  9. Mar 7, 2010 #8
    I couldn't help but notice...

    Shouldn't "-0.51 = vsin22t - 0.5gt^2" be -0.51 = vsin22t + 0.5gt^2?

    The formula is d = Vit + 1/2 at^2 is it not?
     
  10. Mar 9, 2010 #9
    In a problem where the net acceleration is in a direction opposite to the positive direction you chose, the "a" in your formula becomes negative.

    In other words, when you move in the positive direction (in this case upwards) you slow down, when you move in the negative direction (downwards) you speed up.
     
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