/*! 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 52 A movie stuntwoman drops from a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 52

A movie stuntwoman drops from a helicopter that is 30.0 \(\mathrm{m}\) above the ground and moving with a constant velocity whose components are 10.0 \(\mathrm{m} / \mathrm{s}\) upward and 15.0 \(\mathrm{m} / \mathrm{s}\) horizontal and toward the south. You can ignore air resistance. (a) Where on the ground (relative to the position of the helicopter when she drops) should the stuntwoman have placed the foam mats that break her fall? (b) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{j}-t\) graphs of her motion.

Short Answer

Expert verified
The stuntwoman should place the mats 59.1 m south of the drop point.

Step by step solution

01

Understand the motion components

The stuntwoman has two components of motion: horizontal and vertical. The horizontal motion is at a constant velocity of 15.0 m/s toward the south. Vertically, she starts moving upward at 10.0 m/s but is also subject to gravity, acting downward at 9.8 m/s虏.
02

Calculate time of flight vertically

Use the kinematic equation for vertical motion to find the time it takes for the stuntwoman to hit the ground. The equation is: \( y = v_{i_y}t + \frac{1}{2} a t^2 \), where \( y = -30 \text{ m} \) (she falls 30 m), \( v_{i_y} = 10 \text{ m/s} \), and \( a = -9.8 \text{ m/s}^2 \). Solve for \( t \).
03

Solve the quadratic equation

Substitute the values into the quadratic: \[ -30 = 10t - 4.9t^2 \]. Rearrange to \[ 4.9t^2 - 10t - 30 = 0 \]. Solve using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 4.9 \), \( b = -10 \), and \( c = -30 \).
04

Find the positive root for time

Calculating using the quadratic formula gives two roots, but only the positive one is physically meaningful. Solving for time results in approximately \( t = 3.94 \text{ seconds} \).
05

Calculate horizontal distance traveled

The horizontal distance is a result of the constant velocity: \( d = v_{x}t \). Substitute \( v_{x} = 15 \text{ m/s} \) and \( t = 3.94 \text{ s} \). This results in \( d = 15 \times 3.94 \text{ m} = 59.1 \text{ m} \).
06

Draw motion graphs

Draw the following: - **x-t graph**: A straight line with a constant positive slope, since horizontal motion is constant. - **y-t graph**: A parabolic curve starting above the x-axis (at 30 m) moving downward. - **v鈧榵-t graph**: A horizontal line, as horizontal velocity is constant (15 m/s). - **v鈧榶-t graph**: A line starting at 10 m/s and decreasing linearly due to gravity.

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 91影视!

Key Concepts

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

Kinematic Equations
Kinematic equations are essential tools in physics that allow us to predict and analyze motion even without having to observe it directly. These equations relate initial velocity, final velocity, acceleration, time, and displacement. A primary kinematic equation used in vertical motion scenarios is:\[ y = v_{i_y}t + \frac{1}{2} a t^2 \] This equation helps calculate how far an object will travel vertically under gravity, where:
  • \( y \) is the vertical displacement.
  • \( v_{i_y} \) is the initial vertical velocity.
  • \( a \) is the acceleration due to gravity, often approximated as \( 9.8 \text{ m/s}^2 \) downward.
  • \( t \) is the time elapsed.
In our example, the stuntwoman's motion is influenced by these factors. Her initial upward velocity gradually decreases due to gravity, necessitating the solution of a quadratic equation to find her time of flight. Understanding these equations provides clarity on how objects move when influenced by different forces.
Vectors in Physics
Vectors are fundamental in physics to describe quantities that have both magnitude and direction, such as velocity, force, and displacement. Understanding vectors allows us to better describe the multifaceted nature of motion. In the exercise, the stuntwoman's motion consists of two vector components:
  • Horizontal Component: Her horizontal velocity is \( 15.0 \text{ m/s} \) towards the south, which remains constant throughout the fall since no horizontal force like air resistance is acting against it.
  • Vertical Component: Starting with \( 10.0 \text{ m/s} \) upward, this part of her motion is influenced by gravitational acceleration \( 9.8 \text{ m/s}^2 \) downwards.
Combining these components gives a complete picture of her trajectory, making vectors indispensable for solving problems involving two-dimensional motion like ours. Recognizing how vectors operate ensures accurate calculation and prediction of an object's movement path.
Motion Graphs
Graphs are a powerful way to visualize motion, providing an intuitive understanding of the relationships between variables like displacement, velocity, and time. Here are the types of motion graphs relevant for this scenario:
  • x-t graph: Representing horizontal position over time, this graph is a straight line with a positive slope due to the constant southward motion.
  • y-t graph: This graph shows a parabolic path as the stuntwoman rises slightly before falling under gravity, representing the vertical position over time.
  • v鈧榵-t graph: A horizontal line here signifies that horizontal velocity remains consistent over time, highlighting constant motion in the x-direction.
  • v鈧榶-t graph: Depicting negative acceleration, this line starts at \( 10 \text{ m/s} \) and slopes downward as the velocity reduces due to gravity's pull.
Understanding these graphs aids in grasping the full scope of an object鈥檚 journey through space. By breaking down complex equations and results visually, graphs simplify and elucidate the characteristics of motion in a clear and engaging way.

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

The coordinstes of a bird flying in the \(x y\) -plane are given by \(x(t)=\alpha t\) and \(y(t)=3.0 \mathrm{m}-\beta t^{2},\) where \(\alpha=2.4 \mathrm{m} / \mathrm{s}\) and \(\beta=1.2 \mathrm{m} / \mathrm{s}^{2} \cdot\) (a) Sketch the path of the bird between \(t=0\) and \(t=2.0 \mathrm{s}\) . (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t=20 \mathrm{s}\) . (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{s}\) . At this instant, is the bird speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird turning? If so, in what direction?

An elevator is moving upward at a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . A bolt in the elevator ceiling 3.00 \(\mathrm{m}\) above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings ofm the building? (d) According to the observer in part(c), what distance

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a borizontal path 90.0 \(\mathrm{m}\) above the ground and with a speed of \(64.0 \mathrm{m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

Kicking a Field Goal. In U.S. football, after a touchdown the team has the opportunity to earn one more point by kicking the ball over the bar between the goal posts. The bar is 10.0 \(\mathrm{ft}\) above the ground, and the ball is kicked from ground level, 36.0 \(\mathrm{ft}\) horzontally from the bar (Fig. 3.48\() .\) Football regulations are stated in English units, but convert to SI units for this problem. (a) There is a minimum angle above the ground such that if the ball is launched below this angle, it can never clear the bar, no matter how fast it is kicked. What is this angle? (b) If the ball is kicked at \(45.0^{\circ}\) above the horizontal, what must its initial speed be if it to just clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\) .

A rock tied to a rope moves in the \(x y\) -plane. Its coordinates are given as functions of time by $$ x(t)=R \cos \omega t \quad y(t)=R \sin \omega t $$ where \(R\) and \(\omega\) are constants. (a) Show that the rock's distance from the origin is constant and equal to \(R-\) that is, that its path is a circle of radius \(R .(b)\) Show that at every point the rock's velocity is perpendicular to its position vector. (c) Show that the rock's acceleration is always opposite in direction to its position vector and has magnitude \(\omega^{2} R .\) (d) Show that the magnitude of the rock's velocity is constant and equal to \(\omega R\) . (e) Combine the results of parts \((\mathrm{c})\) and \((\mathrm{d})\) to show that the rock's acceleration has constant magnitude \(v^{2} / R .\)
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Space Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Wave Optics

Read Explanation

Force

Read Explanation

Dynamics

Read Explanation
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.