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.)
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
(a) To find the velocity just before the parachute opens, we can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is zero because the parachutist falls freely, and the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2. The time is 10 s.
v = 0 + 9.8 * 10
v = 98 m/s
Therefore, the velocity just before the parachute opens is 98 m/s.
(b) To find the altitude at which the parachute opens, we need to find the distance traveled in the first 10 seconds. We can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Using the given values: u = 0, a = 9.8 m/s^2, and t = 10 s, we can calculate the distance:
s = 0 + (1/2) * 9.8 * (10^2)
s = 490 m
Therefore, the parachute opens at an altitude of 490 m.
(c) To find the velocity just before the parachutist strikes the ground, we need to calculate the final velocity during the time when the parachutist is slowing down with an acceleration of -4.5 m/s^2. We can again use the equation of motion:
v = u + at
In this case, the initial velocity is the velocity just before the parachute opens, which is 98 m/s, the acceleration is -4.5 m/s^2, and the time is 20 s.
v = 98 + (-4.5) * 20
v = 98 - 90
v = 8 m/s
Therefore, the velocity just before the parachutist strikes the ground is 8 m/s.
(d) To calculate the total time required for the whole descent, we can add up the times for each phase of the descent:
Phase 1: Free fall for 10 seconds
Phase 2: Slowing down with an acceleration of -4.5 m/s^2 for 20 seconds
Phase 3: Falling at a uniform velocity for the remaining time
Phase 1 time: 10 s
Phase 2 time: 20 s
Phase 3 time: Total time - Phase 1 time - Phase 2 time
Phase 3 time: unknown - 10 s - 20 s
To find the total time required, we need to calculate the remaining time. We know that the total time is 30 seconds (10 seconds in phase 1 + 20 seconds in phase 2).
Phase 3 time: 30 s - 10 s - 20 s
Phase 3 time: 0 s
Therefore, the total time required for the whole descent is 30 seconds.
(e) To find the height from which the parachutist would have to fall freely in order to strike the ground with the same velocity as she does when wearing a parachute, we need to find the distance traveled in the third phase of the descent (falling at a uniform velocity).
The distance traveled in phase 3 is equal to the velocity (8 m/s) multiplied by the time taken in phase 3, which is the remaining time from part (d).
Distance traveled in phase 3 = 8 m/s * 0 s
Distance traveled in phase 3 = 0 m
Therefore, the parachutist would have to fall from a height of 0 m in order to strike the ground with the same velocity as she does when wearing a parachute.
To find the reasoning behind each answer, let's break down the problem and solve it step-by-step:
Given information:
Initial height, h = 3.1 × 10^3 m
Time when falling freely, t1 = 10 s
Time when slowing down with an acceleration, t2 = 20 s
Acceleration during slowing down, a = -4.5 m/s^2
(a) Velocity just before the parachute opens:
To find the velocity just before the parachute opens, we need to calculate the velocity of the parachutist during the free fall. The formula to calculate the final velocity during free fall is:
vf = vi + g * t
Where:
vf is the final velocity
vi is the initial velocity (which is 0 in this case since the parachutist starts from rest)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time during free fall (t1 = 10 s)
Substituting the values into the formula:
vf = 0 + 9.8 * 10
vf = 98 m/s
So, her velocity just before the parachute opens is 98 m/s.
(b) Altitude where the parachute opens:
To find the altitude where the parachute opens, we need to calculate the distance covered during free fall. The formula to calculate distance during free fall is:
d = (1/2) * g * t^2
Where:
d is the distance covered
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time during free fall (t1 = 10 s)
Substituting the values into the formula:
d = (1/2) * 9.8 * (10^2)
d = 490 m
Therefore, the altitude where the parachute opens is 490 m.
(c) Velocity just before striking the ground:
To find the velocity just before striking the ground, we need to calculate the final velocity during the deceleration phase. The formula to calculate the final velocity during deceleration is:
vf = vi + a * t
Where:
vf is the final velocity
vi is the initial velocity (which is the velocity just before the parachute opens)
a is the acceleration during the deceleration phase (given as -4.5 m/s^2)
t is the time of deceleration (t2 = 20 s)
Substituting the values into the formula:
vf = 98 - 4.5 * 20
vf = 8 m/s
So, her velocity just before striking the ground is 8 m/s.
(d) Time required for the whole descent:
To calculate the time required for the whole descent, we need to find the total time including the free fall time (t1), deceleration time (t2), and the time during uniform velocity.
Total time = t1 + t2 + t_uniform
Given that the total time is 30 seconds (t1 + t2 = 30 s) and the deceleration time is t2 = 20 s, we can solve for the time during uniform velocity:
t_uniform = total time - deceleration time
t_uniform = 30 - 20
t_uniform = 10 s
Therefore, the time required for the whole descent is 30 seconds.
(e) Height to fall freely to achieve the same velocity as with the parachute:
To calculate the height from which the parachutist would have to fall freely to have the same velocity as when wearing a parachute, we need to equate the final velocities in both scenarios.
Velocity during the free fall, vf1 = 98 m/s (from part a)
Velocity during the fall with parachute, vf2 = 8 m/s (from part c)
Using the formula for free fall:
vf1 = sqrt(2gd1)
Where:
d1 is the height for free fall
Squaring both sides of the formula and rearranging it:
(98^2) = 2 * 9.8 * d1
d1 = (98^2) / (2 * 9.8)
d1 = 4900 m
Therefore, she would have to fall freely from a height of 4900 m to achieve the same velocity as she does when wearing a parachute.
By solving these steps, you should have a better understanding of the reasoning behind each answer and how to approach similar problems.