physics - HELP
posted by Anonymous on .
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?
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.