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A model rocket blasts off and moves upward with an acceleration of 12 m/s^2 until it reaches a height of 28 m , at which point its engine shuts off and it continues its flight in free fall.

What is the speed of the rocket just before it hits the ground?

  • Math -

    So there are two steps since two accelerations are involved.

    Step 1: acceleration = +12 m/s/s
    initial velocity, u=0
    final velocity = v
    distance travelled, S=28 m
    using
    V²-u²=2aS
    we find

    =2aS+u²
    =2*12*28+0
    =4√42 m/s (upwards)
    =25.92 m/s (upwards)

    Step 2: free fall, a=-9.81 m/s/s
    initial velocity, u = 4√42 m/s
    distance travelled, S = -28 m
    final velocity = v
    Again, we use the formula
    V²-u²=2aS
    from which

    =2aS+u²
    =2*(-9.81)*(-28)+(4√42)²
    =1221.36
    v=√1221.36
    =34.95 m/s (downwards)

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