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Simple projectile motion

  1. Oct 11, 2005 #1
    I've done these questions a million times, but it was last year and I'm stumped. if you have:

    velocity of projectile
    angle
    range

    and the initial height is greater than that of the final height, how do you solve for initial height and final velocity?
     
    Last edited: Oct 11, 2005
  2. jcsd
  3. Oct 11, 2005 #2
    the law of conservation of energy is much better :P
     
  4. Oct 30, 2008 #3
    I have a similar issue.. I took AP Phys B last year and this year I can't remember the simplest things!
    I am completely stumped on a problem whose only known variables are the maximum height of the trajectory and the horizontal distance the object travels.
    I can't seem to figure out a way to find the velocities involved!
    Help...
     
  5. Oct 30, 2008 #4
    Perhaps giving us exact examples would help, also showing what you have tried is always a good start.
     
  6. Oct 30, 2008 #5
    The problem is really simple. It involves a guy throwing a pebble at a window such that when it hits the window at a height of 4.5 m, it has only horizontal velocity (it has reaches the top of its trajectory). The distance to the window horizontally is 5.0 m, and I am asked for the horizontal velocity.
    No matter which equation I try, I always seem to have more than one unknown; for example, the position equation x=initial position + v(initial)t + .5at^2. I'd need v to solve for t and vice versa.
    The only thing that I thought might work was neglecting the window and assuming a complete trajectory, but I still end up with more than one variable.
    the more I look at this problem, the more muddled I become :(
     
  7. Oct 30, 2008 #6
    I also tried calculating the angle of the trajectory based on the distances given and came up with approximately 41 degrees... but from there I didn't know how to calculate the vector components of the velocity because no velocities are given.
     
  8. Oct 30, 2008 #7
    So you're solving for the height it's thrown from correct? You don't have final velocity or initial velocity? Well if you ask me, you're in a bit of a pickle..
     
  9. Oct 30, 2008 #8
    no problem, thanks for your attention to this!
     
  10. Oct 30, 2008 #9
    no, I'm solving for the horizontal velocity of the pebble. I know the height of the trajectory, so the height from which it is thrown is irrelevant, I think. I am also given the horizontal distance to the window.
    Basically, my issue is that I can't figure out how to isolate and solve for any of the velocities.
     
  11. Oct 30, 2008 #10
    What do you mean you know the height of the trajectory..? The trajectory under uniform, one dimensional, constant acceleration is parabolic. The trajectory is not a constant value.

    If you knew the height it was thrown from, we could solve it.
     
  12. Oct 30, 2008 #11
    sorry, I meant the maximum height that the pebble reaches relative to the point from which it was thrown..
     
  13. Oct 30, 2008 #12
    Okay, now we're getting somewhere. How far (horizontally and vertically) is this from its final point?
     
  14. Oct 30, 2008 #13
    It is 5.0 m from the window horizontally, and 4.5 m away vertically.
     
  15. Oct 30, 2008 #14
    Okay, this is good. I now realise that you mentioned this earlier. My apologies for not picking up on it then.

    First we want to find the time that it takes for the projectile to fall from this peak point to the same vertical height as the window (we'll label this point zero).
    From this, we have:

    Acceleration = 9.8ms-2
    Displacement(y) = 4.5m
    initial velocity (y) = 0ms-1
    you can now solve for the time it takes to fall the distance, I'll give you a head start.

    [tex] s = ut + \frac{1}{2}at^2 [/tex]

    where u = initial velocity
    t = time
    a = acceleration
    s = displacement

    Let me know if you're understanding this (I don't have a good gauge of how advanced you are yet)
     
  16. Oct 30, 2008 #15
    I understand this explanation, but I fear I've miscommunicated the problem :(
    4.5 meters is both the maximum height that the pebble reaches as well as the height at which it hits the window.
     
  17. Oct 30, 2008 #16
    Okay, that makes me very confused then. When it is 5m horizontally from the window, it is 4.5m vertically from the ground. And when it is 0m horizontally from the window (when it hits it) it's is also 4.5m vertically from the window.

    And yet 4.5 is the highest point in the pebbles trajectory? This doesn't really make sense to me sorry. Is there a diagram we can refer to?
     
  18. Oct 30, 2008 #17
    let's see... there's no diagram that I know of..
    I'm assuming the point from which it was thrown as "ground level" or zero vertical distance. I figure the height of the man throwing the pebble is irrelevant to its motion.
    That way, the pebble hits the window at 4.5 m from its starting point at zero, and the horizontal distance remains 5.0 m.
     
  19. Oct 30, 2008 #18
    [​IMG]

    Okay I hope we've got this right this time!

    So that's all the information we're given, are we specifically told that it hits the window at the peak of its trajectory?

    We can also assume that gravity is 9.8ms-2

    If all of this is true, then we can solve the equation.
     
  20. Oct 30, 2008 #19
    yes! I think we've got it this time.
    Thank you so much for spending this time!
     
  21. Oct 30, 2008 #20
    Just realised that I got 5m and 4.5 m backwards haha.

    Okay so first we work with vertical axis.
    We have:

    Acceleration (9.8ms-2)
    Final velocity (0ms-1)
    displacement (4.5m)
    solving for initial velocity (am I right?)

    so first for y velocity:

    [tex] v^2 = u^2 + 2as [/tex]
    where v = final velocity
    u = initial velocity
    a = acceleration
    s = displacement.

    Let me know how you go with this.
     
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