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March 28, 2017

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A parachustist jumps from an airplane and freely falls y=51.8 m before opening his parachute. Thereafter, he decelerates at a=1.86 m/s^2. As he reaches the ground, his speed is 3.04 m/s.

a)How long was the parachutist in the air?

b)At what height did the parachutist jump from his plane?

  • physics - HELP - ,

    There are two parts to the motion:
    1. free fall over a distance of 51.8 m.
    initial velocity, u=0 m/s
    final velocity, v to be determined.
    We ignore air resistance for this part.
    acceleration due to gravity, a = -g = -9.8 m/s/s
    Using the formula
    v²-u² = 2aS
    v can be determined.
    (see answer to next question if necessary).
    Time t1 = (v-u)/a

    2. Parachute open (in infinitesimal time).
    initial velocity, v (determined from part 1) m/s
    final velocity,w = 3.04 m/s
    acceleration, a = -1.86 m/s/s
    Time, t2 = (w-v)/a
    The distance descended when parachute is open can be determined also by:
    w²-v² = 2aS

    Total time = t1+t2

    Add up distance for parts 1 and 2 will give the total distance.

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