Posted by **Anonymous** on Saturday, September 19, 2009 at 9:10pm.

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 -
**MathMate**, Saturday, September 19, 2009 at 9:38pm
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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