Posted by jenny on .
A parachutist bails out and freely falls 50m. Then the parachute opens and thereafter she decelerates at 2 m/s^2. She reaches the ground with a speed of 3m/s.
a) how long is the parachutist in the air?
b)At what height does the fall begin?
for a) would I use x-x0 =vt-1/2 at^2 and find t? where x-x0= 50m, v= 3m/s, a = 9.8m/s^2
I am not sure what equation to use for b)
I would answer this in 2 parts. First, I would calculate the time she is in the air BEFORE she opens the chute...using x = v0t - 1/2 at^2. v0 here is 0, so you can calculate time. Then I would calculate the final velocity WHEN she opens the chute using
vf = v0 + at. v0 is 0, and you now know time t.
Then, knowing the velocity when the chute opens, you can calulate the time using vf = v0 - 2t.
To get the total time, just add the 2 times you got.
To get the total height...well, you already know the first part...50m.
To get the second height, use x = v0t-1/2 at^2, and add the two.
I hope this explains it. If not, repost!