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Cannon fired at moving car

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data
    A cannon is fired(from pt. A) at an angle of 38deg and with a Vo of 1000 ft/sec. A car leaves (from pt. C) at 30 mph heading towards the cannon. The cannon hits the truck at point B.
    a. Determine the horizontal distance R that the cannonball travels to hit the car.
    b. How long did it take to hit the car?
    c. How far did it travel from point C before it was hit?
    d. If length of the car is 10 ft. How long does the car have when it can be hit?

    2. Relevant equations
    X= (Vo^2(sin2theta))/g
    X= Xo + (Vo)t
    X=Xo +Vot +1/2(Ac)(t^2)

    3. The attempt at a solution
    a. Using the first equation, plugging in the initial velocity(1000ft/sec), angle(38) and dividing by gravity(32.152ft/s^2) yields a solution of 30175ft.

    b. plugging the distance travelled into the second equation to determine t yields a solution of 38.29s

    c. Since the cannon hits the car after 38.29s, then the car must have travelled for that same length of time, initially leaving at 30mph.
    Would it be as simple as x=vt to determine the distance? I come up with 1684 ft(after converting from mph to ft/sec) Or is there some way to use simultaneous equations to find Ac, then determine the distance travelled?

    d. Not sure how to begin this one, but if the car is 10 ft long, then the time it takes to move 10 ft would seem to be the time when it could be hit.
  2. jcsd
  3. Mar 9, 2008 #2


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    Science Advisor
    Homework Helper

    Hi jt! Welcome to PF! :smile:

    I'm not sure what you've done, but it looks suspicious …

    The cannonball's horizontal velocity is constant, and is V0.cos38º.

    The cannonball's initial vertical velocity is V0.sin38º, and you have to apply the usual equations to find out the time t when that velocity is reduced to zero.

    Then the cannonball returns to the road, and hits the car, at time 2t. :smile:
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