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Projectile Motion ~ AP Physics (A rifle is fired at an angle of )

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  1. Sep 17, 2012 #1
    A rifle is fired at an angle of 25° below the horizontal with an initial velocity of 400 m/s from the top of a cliff 60.0m high. How far from the base of the cliff does the bullet land?

    This is what I did, but i'm not sure if its right:

    y-component→ v=400sin(25°) ∴v=169.05
    x-component→ v=400cos(25°)∴v=362.52
    I found the time by doing: t= 60/169.05∴t= .355s
    From here I multiplied the time with the x-component to find the distance(X) : X=.355(362.52)∴ X=128.69m


    Is this right???
    If not please help me.
     
  2. jcsd
  3. Sep 17, 2012 #2
    don't forget about the constant acceleration due to gravity in the y-direction
     
  4. Sep 17, 2012 #3
    Could you explain a bit more, @SISHKABOB
     
  5. Sep 17, 2012 #4
    what I mean is that the time that you determined is the time that it would take an object to travel 60m while going a constant 169.05 m/s

    but in the y direction, there is a constant acceleration due to gravity, so the projectile will be speeding up as it falls. It will start at 169.05 m/s, but will accelerate until it hits the ground.

    You need to take this into account.
     
  6. Sep 17, 2012 #5
    @SISHKABOB, So you mean I should use this equation: Y(distance)=vt+.5at²
    so that would be: 60=(169.05)t+.5(-9.8)t² therefore t=.359s
     
  7. Sep 17, 2012 #6
    yep. It's a very small difference but that's only because the problem is talking about a bullet shot from a rifle. In slower situations it makes a much bigger difference.

    try setting the initial velocity to like, 1 m/s, and then find the difference in time if you did it first ignoring gravity, and then considering gravity
     
  8. Sep 17, 2012 #7
    I see what you mean. Thank you!
     
  9. Sep 17, 2012 #8

    the bullet is shot downwards, and the acceleration from gravity is also downwards, so these should have the same sign. If the top of the cliff is at 0, the bottom should be at -60, and you should get -60 = (-169.05)t + (0.5) (-9.8)t² => 60=(169.05)t+.5(9.8)t

    and the bullet should hit the ground a little faster than in the unaccelerated case, instead of slower.
     
  10. Sep 17, 2012 #9
    oops! you're absolutely right, I wasn't thinking it through
     
  11. Sep 17, 2012 #10

    K^2

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    When you add an extra effect, like gravity, it's always good to compare your answer with answer that neglected said effect. Fact that time came out longer should have been an immediate warning sign.

    Of course, in this problem, effect of gravity is almost negligible. Velocity achieved by a bullet in free-fall from 60m is about 35m/s, which is far less than initial velocity of 170m/s. That's how you tell that you probably don't need to worry about it for any practical purpose. For an AP Physics question, however, they probably expect you to factor in gravity.
     
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