/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 A parachutist bails out and free... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A parachutist bails out and freely falls \(50 \mathrm{~m}\). Then the parachute opens, and thereafter she decelerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). She reaches the ground with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). (a) How long is the parachutist in the air? (b) At what height does the fall begin?

Short Answer

Expert verified
The parachutist is in the air for 17.32 seconds, and the fall begins at 292.11 meters.

Step by step solution

01

Calculate the Time of Free Fall

First, calculate the time taken for the parachutist to fall freely for 50 m. Use the equation for free fall: \[ s = \frac{1}{2}gt^2 \] where \( s = 50 \) m is the distance and \( g = 9.8 \) m/s² is the acceleration due to gravity. Solving for \( t \): \[ 50 = \frac{1}{2} \times 9.8 \times t^2 \] \[ 100 = 9.8t^2 \] \[ t^2 = \frac{100}{9.8} \] \[ t \approx 3.19 \text{ seconds.} \]
02

Calculate the Velocity After Free Fall

Next, find the velocity of the parachutist just before the parachute opens using the formula: \[ v = gt \] where \( t \approx 3.19 \) s is the time during free fall. \[ v = 9.8 \times 3.19 = 31.26 \text{ m/s.} \]
03

Calculate Deceleration Time

Calculate the time taken to decelerate from the velocity when the parachute opens (31.26 m/s) to the final velocity (3.0 m/s) using \( v = u + at \) where \( a = -2.0 \) m/s² is the deceleration: \[ 3 = 31.26 + (-2)t \] \[ -28.26 = -2t \] \[ t = \frac{28.26}{2} = 14.13 \text{ seconds.} \]
04

Calculate Total Time in Air

Add the time of free fall to the deceleration time to find the total time in the air: \[ t_{total} = 3.19 + 14.13 = 17.32 \text{ seconds.} \]
05

Calculate the Height from Which the Parachutist Fell

Use the total time and deceleration to find the distance traveled after the parachute opens. Use \( s = ut + \frac{1}{2}at^2 \) with \( u = 31.26 \text{ m/s}, a = -2.0 \text{ m/s}^2, t = 14.13 \text{ s} \): \[ s = 31.26 \times 14.13 + \frac{1}{2}(-2)(14.13)^2 \] \[ s \approx 441.55 - 199.44 = 242.11 \text{ m.} \] Add the distance of free fall (50 m) to this distance to find the total height: \[ h = 50 + 242.11 = 292.11 \text{ m.} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Fall
Free fall is an essential concept in physics that describes the motion of objects solely under the influence of gravity.During free fall, objects experience a uniform acceleration due to Earth's gravitational pull, typically approximated as 9.8 m/s².In the case of the parachutist, free fall begins once she bails out. She falls freely for a distance of 50 meters.The equation used to calculate the time of free fall is:\[ s = \frac{1}{2}gt^2 \]Here, \( s \) represents the distance fallen, and \( g \) is the acceleration due to gravity.
The equation is solved for \( t \), the time taken to fall, providing insight into how long the parachutist is in free fall.Understanding free fall helps us predict the motion and timing of falling objects without the interference of other forces like air resistance.
Deceleration
Deceleration is a term used to describe the reduction in velocity of an object.In kinematics, it is essentially the opposite of acceleration. For a parachutist, once the parachute is deployed, the concept of deceleration becomes vital.As the parachute opens, it creates air resistance, causing the parachutist to slow down progressively.This slowing down is quantified as a deceleration rate, in this case, \[ a = -2.0 \, \mathrm{m/s^2} \] This means the parachutist reduces her speed by 2 meters per second every second until reaching a safer landing velocity of 3 meters per second.Understanding deceleration is crucial for ensuring a safe descent, especially in activities involving high initial speeds.
Parachute Descent
A parachute descent represents the combination of physics principles to safely bring a parachutist to the ground.
The descent begins when the parachute opens, marking the transition from free fall to controlled slow fall. The parachute's role is to increase air resistance, effectively reducing the parachutist's speed. After initial free fall, the parachutist undergoes deceleration, with the parachute creating a significant opposing force.
This force is higher than the gravitational pull, ensuring a gentle reduction in speed.
Eventually, the parachutist reaches a steady velocity, known as terminal velocity, before landing.
The final descent distance is calculated using mechanics equations, helping determine the total height from which a parachutist falls. Understanding parachute descent is essential for correctly predicting both the time and position during the entire jump.
Velocity Equations
Velocity equations are pivotal in solving problems related to motion and speed changes.For the parachutist scenario, velocity equations help compute various stages of her jump.
Before the parachute opens, the velocity is calculated at the end of the free fall using:\[ v = gt \]This provides the speed just before the parachute deploys.For deceleration, another basic velocity equation is used:\[ v = u + at \]Here, \( u \) is the initial velocity post-free fall, \( a \) is the deceleration, and \( t \) is time.
This equation is crucial to find how speed transitions from free fall to a slower speed upon parachute deployment.Understanding and applying these equations allow for the complete analysis and safe planning of parachute jumps.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A bolt is dropped from a bridge under construction, falling \(90 \mathrm{~m}\) to the valley below the bridge. (a) In how much time does it pass through the last \(20 \%\) of its fall? What is its speed (b) when it begins that last \(20 \%\) of its fall and (c) when it reaches the valley beneath the bridge?

An automobile travels on a straight road for 40 \(\mathrm{km}\) at \(30 \mathrm{~km} / \mathrm{h}\). It then continues in the same direction for another \(40 \mathrm{~km}\) at \(60 \mathrm{~km} / \mathrm{h}\). (a) What is the average velocity of the car during the full \(80 \mathrm{~km}\) trip? (Assume that it moves in the positive \(x\) direc-

A red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an \(x\) axis, At time \(t=0\), the red car is at \(x_{r}=0\) and the green car is at \(x_{g}=\) \(220 \mathrm{~m}\). If the red car has a constant velocity of \(20 \mathrm{~km} / \mathrm{h}\), the cars pass each other at \(x=44.5 \mathrm{~m}\), and if it has a constant velocity of \(40 \mathrm{~km} / \mathrm{h}\), they pass each other at \(x=76.6 \mathrm{~m}\). What are (a) the initial velocity and (b) the constant acceleration of the green car?

An electron has a constant acceleration of \(+3.2 \mathrm{~m} / \mathrm{s}^{2}\). At a certain instant its velocity is \(+9.6 \mathrm{~m} / \mathrm{s}\). What is its velocity (a) \(2.5 \mathrm{~s}\) earlier and (b) \(2.5\) s later?

An electron with an initial velocity \(v_{0}=1.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) enters a region of length \(L=1.00\) \(\mathrm{cm}\) where it is electrically accelerated (Fig. 2-23). It emerges with \(v=5.70 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is its ac- celeration, assumed constant?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.