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Final Velocity with barely anything AP Phys help

  1. Sep 18, 2004 #1
    A boy fires a "BB" gun at an angle of 28o above the horizontal. The "BB" strikes a target that is a vertical distance of 1.6 meters above and 12.8 meters to the right of the muzzle (exit point) of the gun. Assuming the bullet has a negative vertical velocity when it strikes the target, calculate the bullet's velocity when it leaves the gun.

    i tried everything that i could think of, using trig to figure out

    Vx=Vy/tan(28) and therefore Vy/tan(28)=12.8

    anything that i would plug into the y=VoT + 1/2(a)(Tsqrd) i would get a negative square root

    Thanks in Advance
  2. jcsd
  3. Sep 18, 2004 #2
    Welcome to PF Anshu...

    Well the obvious line of attack for such a problem would be to write the expressions for the horizontal and vertical distances in terms of time and eliminate time to get a trajectory equation of the form,

    [tex]y - y_{0} = (x-x_{0})\tan \alpha -\frac{1}{2}\frac{g(x-x_{0})^{2}}{V_{0}^2}\sec^2\alpha[/tex]

    and then plug in all that you have to solve for the initial velocity. (First, try and derive the above equation to convince yourself that it is true).

    The other method is to solve for Vx and Vy somehow and relate them through the angle they make. My question here: What do you think the angle between Vx and Vy will be at the time the shot hits the wall? Will it be 28 degrees? Are you sure of that?

    Also, you have mentioned an equation for y, the vertical distance. What do you propose to do with it?

    Hope that helps...

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