a) v = at
where a is gravity
b) y = 1/2 g t^2
chute opens at 3.1e3 - y
c) vf = vi + at
vi is what you found in a)
a = -4.5
t = 20
d)time while slowing down
vf = vi +at
solve for t
now we need the distance she fell while slowing
y = vot + 1/2at^2
3.1e3 - y(b) - y= what's left
t = that distance over final velocity
add up the times
e) y = v^2/2g
I want to know the reasoning behind each answer as well cause I have the answers, I want to know why and how to do the answers as well :)
A parachutist jumps from a height of 3.1 × 10[Power 3] m and falls freely for 10 s. She then opens her parachute, and for the next 20 s slows down with an acceleration of –4.5 m/s[Power 2] . After that, she falls the rest of the distance to the ground at a uniform velocity.
(a) What is her velocity just before the parachute opens?(b) At what altitude does the parachute open?(c) What is the velocity of the parachutist, just before she strikes the ground?(d) Calculate the time required for the whole descent.(e) From what height would she have to fall freely in order to strike the ground with the same velocity as she does when wearing a parachute? (This is how parachutists are trained.)
1 answer