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Projectile Motion

  1. Mar 14, 2006 #1
    A rocket is launched at an angle of 53o above the horizontal with an initial speed of 100m/s. It moves for 3s along its initial line of motion with an accelration of 30m/s2. At this time its engines fail and the rocket proceeds to move as a free body. Find a)the maximum altitude reached by the rocket, b)its total time of flight, and c)its horizontal range.

    Can someone please explain how to approach this problem.
  2. jcsd
  3. Mar 14, 2006 #2


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    Start by figuring out its velocity after 3 seconds of accelerating.
  4. Mar 15, 2006 #3
    Need More Help Please.

    This is what i have so far.

    Part a:
    s = vot + (.5)at^2
    s = 100m/s(3s) + .5(30m/s/s)(3s)^2 = 435 m

    y1 = s(sin 53) = 347 m
    y2= [100(sin53)^2 / 2(9.8)] = 325 m
    ytotal = 347 + 325 = 672 m

    correct answer is 1520 m

    The rest of the problem will be solved incorrectly. But here is what i would have done if the previous answer was correct.

    Part b:
    t(total) = t1 + t2

    t1= 3s
    y= vo(sin53)t - .5*g*(t^2)
    solve for time using quadratic eqn.

    the correct answer is 36.1 s

    Part c
    x1 = v0t + .5(a)t^2 = 262m
    x2 = s(cos 53) = 435m
    xtotal = 435 + 262 = 697 m

    the correct answer is 4050 m
  5. Mar 15, 2006 #4


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    You've already figured out that it is 347 meters above the ground after you finish accelerating. Now you need to know your y-component velocity after you've finished accelerating. With that, you can compute how long it will take gravity to slow you to 0, and using your velocity after accelerating and 0, you can compute an average velocity. With your average velocity and time, you can compute how high the rocket will climb after it finishes accelearating, and add this to your 347 meters for your final height.
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