Kinematics Question

Question

  A parachutist jumps from a height of 3.1 x10³ m and free-falls for 10 s. She then opens her parachute and for the next 20 s slows down with an acceleration of -4.5 m/s². 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 her velocity just before she lands?

d. How long does the entire jump last?

e. From what height would she need to jump from (without a parachute) in order to hit the ground at the same speed as in c)?

Answer 1

Picking downwards motion as positive, a subscript is when chute opens, c subscript is when chute stops decelerating...

a) v_a =gt

b) distance fallen, d_a = 1/2gt² where t is 10 seconds, altitude is h - d_a = h_a

c) The velocity of landing is the same as velocity achieved 20 seconds after parachute opens:

v_c = v_a - 4.5 m/s² (20 seconds)

d) t for the jump is 30 seconds + the time to fall at constant velocity. First we need the height when the deceleration is complete. d_c = 1/2 (v_a + v_c) (20 s) and h_c = h_a - d_c

Final time = 30 s + h_c/v_c

e) h = (v_c²)/(2g) from 2ad = v_f² - v_i²

Answer 2

a) use (final velocity) = (initial velocity) + acceleration*time

initial velocity = 0 m/s because they are starting from rest

acceleration = -9.8 m/s^2 from gravity

time = 10 seconds

plugging in (final velocity) = 0m/s + (-9.8m/s^2*10s) = -98m/s

b) use (final position) = (initial position) + 0.5*(initial velocity + final velocity)*time

initial position = 3.1*10^3 m

initial velocity = 0 m/s

final velocity = -98 m/s (from part a)

time = 10 s

plugging in (final position) = 3.1*10^3 m + 0.5*(0 m/s -98m/s)*10s = 2.61*10^3 m

c) then we have to determine what the velocity is after the 20 seconds of decelerating at 4.5 m/s^2

use (final velocity) = (initial velocity) + acceleration*time

initial velocity = -98 m/s (from part b)

acceleration = 4.5 m/s^2 (positive because acceleration is upwards because of air drag of parachute)

time = 20 sec

plugging in: (final velocity) = -98 m/s + 4.5 m/s^2*20 sec = -8 m/s

since the velocity is uniform the rest of the way down the velocity is -8 m/s just before landing

d) so far we have solved for the first 30 seconds of the jump. now we need to find out how long it takes to get to the ground at -8 m/s

first determine distance covered when parachute is out

use (final position) = (initial position) + 0.5*(initial velocity + final velocity)*time

initial position = 2.61*10^3 m

initial velocity = -98 m/s

final velocity = -8 m/s

time = 20 sec

plugging in (final position) = 2.61*10^3 m + 0.5*(-98m/s -8m/s)*20sec = 1.55*10^3 meters

so now we know that the parachutist has 1.55*10^3 meters left to fall at a velocity of -8 m/s so solving for time: 1.55*10^3 m/8m/s = 193.75 seconds

so the entire jump lasts 20 sec + 193.75 sec = 213.75 sec

e) use EQ1:(final position) = (initial position) + 0.5*(initial velocity + final velocity)*time and

EQ2: (final velocity) = (initial velocity) + acceleration*time

where final position = 0

initial position = what we are solving for

initial velocity = 0 m/s

final velocity = -8 m/s

acceleration = -9.8 m/s

rearrange EQ2: time = (final velocity -initial velocity)/acceleration

plug time from EQ2 into EQ1:

(final position) = (initial position) + 0.5*(initial velocity + final velocity)*(final velocity -initial velocity)/acceleration

plug in values solving for initial position

initial position = (final position) - 0.5*(initial velocity + final velocity)*(final velocity -initial velocity)/acceleration = 0 - 0.5*(0m/s-8m/s)*(-8m/s-0m/s)/(-9.8m/s^2)

initial position = 3.26 m