/*! 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 72 Tossing Your Lunch. Henrietta is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 72

Tossing Your Lunch. Henrietta is going off to her physics class, jogging down the sidewalk at 3.05 \(\mathrm{m} / \mathrm{s} .\) Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is 38.0 \(\mathrm{m}\) above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally 9.00 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. (a) With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground? (b) Where is Henrietta when she catches the bagels?

Short Answer

Expert verified
Bruce throws at 9.87 m/s; Henrietta is 35.93 m away.

Step by step solution

01

Determine Henrietta's Position After 9 Seconds

Henrietta is jogging at a constant speed of \(3.05\, \mathrm{m/s}\). To find how far she has traveled in 9 seconds, use the formula for constant velocity: \(\text{distance} = \text{speed} \times \text{time}\). Thus, \(\text{distance} = 3.05\, \mathrm{m/s} \times 9\, \mathrm{s} = 27.45\, \mathrm{m}\). Therefore, Henrietta is 27.45 meters away from the point directly under the window when Bruce throws the bagels.
02

Calculate the Time for Bagels to Fall

The bagels are thrown horizontally from a height of 38.0 meters. We need to find the time it takes for them to hit the ground using the vertical falling motion equation: \[ y = \frac{1}{2} g t^2 \] where \(y = 38.0\, \mathrm{m}\) and \(g = 9.8\, \mathrm{m/s^2}\).Solving for \(t\), \( 38.0 = \frac{1}{2} \times 9.8 \times t^2 \), gives \( t^2 = \frac{38.0 \times 2}{9.8} \approx 7.755 \). Therefore, \( t \approx \sqrt{7.755} \approx 2.78 \) seconds.
03

Determine the Initial Speed of the Bagels

Bruce must throw the bagels such that they cover a horizontal distance of 27.45 meters in the same time they are falling. Use the formula for horizontal distance: \( x = v_x t \) where \(x = 27.45\, \mathrm{m}\) and \(t = 2.78\, \mathrm{s}\) from Step 2.Solving for \(v_x\), \( v_x = \frac{x}{t} = \frac{27.45}{2.78} \approx 9.87\, \mathrm{m/s} \).Thus, Bruce must throw the bagels with an initial speed of approximately \(9.87\, \mathrm{m/s}\).
04

Find Henrietta's Position When Catching the Bagels

Henrietta continues to run at 3.05 m/s for the duration of the bagels' fall, which is 2.78 seconds, after Bruce throws them. Calculate her additional movement:\(\text{additional distance} = 3.05\, \mathrm{m/s} \times 2.78\, \mathrm{s} \approx 8.48\, \mathrm{m}\).Adding this to her initial distance from the window (27.45 m), Henrietta is \(27.45 + 8.48 = 35.93\, \mathrm{m}\) away from the window when she catches the bagels.

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.

Horizontal Velocity
In projectile motion, the horizontal velocity is crucial in determining the distance an object will travel horizontally. Horizontal velocity is unique because it remains constant if we ignore forces like air resistance. Unlike vertical motion, there is no acceleration acting horizontally, so the velocity stays the same from the moment the object is thrown or released. This is why, when Bruce throws the bagels horizontally, they maintain the same speed throughout their flight.

For example, if you throw a ball horizontally off a cliff, its horizontal velocity won't change whether the ball is high up or closer to the ground. This constant horizontal velocity is calculated using the equation:
  • Horizontal Distance: \( x = v_x \, t \)
Where \( v_x \) is the horizontal velocity, and \( t \) is the time of flight. In this exercise, because Herniietta continues to jog, the horizontal velocity must be calculated precisely, using the time and distance she covers while the bagels are in the air.
Vertical Motion
Vertical motion in projectile problems involves an object's movement under the influence of gravity. Unlike horizontal motion, vertical motion experiences a constant acceleration due to gravity. This acceleration is approximately \( 9.8 \, \mathrm{m/s^2} \) towards the Earth. When solving for the time it takes for the bagels to hit the ground, we use this information.

The vertical distance the bagels need to fall is calculated using the formula:
  • Vertical Distance: \( y = \frac{1}{2} g t^2 \).
By plugging in the height (38 meters) and solving for the time \( t \), you can determine how long the bagels will be in the air. Remember, no matter how fast or slow the object is moving horizontally, the time to fall from a certain height remains the same.

This constancy helps to coordinate the timing of Henrietta's catch since the time it takes the bagels to fall controls the entire event's timing.
Constant Velocity
Constant velocity refers to motion where the speed and direction remain unchanged. In Henrietta's case, she jogs at a constant speed of \( 3.05 \, \mathrm{m/s} \), meaning her velocity doesn't vary over the time she runs. This concept simplifies the calculation of her position, as we can simply multiply her velocity by the time to find out how far she has run.

This is shown in the equation:
  • Position: \( \text{distance} = \text{speed} \times \text{time} \)
Because of her constant speed, it is straightforward to compute exactly where she is when Bruce throws the bagels and when she catches them. Her constant speed ensures that there are no surprises in her position, making it possible to precisely time her catch with the bagels thrown from above. Understanding constant velocity helps in accurately predicting and solving many real-world physics problems, including this one.

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

When it is 145 \(\mathrm{m}\) above the ground, a rocket traveling vertically upward at a constant 8.50 \(\mathrm{m} / \mathrm{s}\) relative to the ground launches a secondary rocket at a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(53.0^{\circ}\) above the horizontal, both quantities being measured by an astronaut sitting in the rocket. After it is launched the secondary rocket is in free-fall. (a) Just as the secondary rocket is launched, what are the horizontal and vertical components of its velocity relative to (i) the astronaut sitting in the rocket and (ii) Mission Control on the ground? (b) Find the initial speed and launch angle of the secondary rocket as measured by Mission Control. (c) What maximum height above the ground does the secondary rocket reach?

A student sits atop a platform a distance \(h\) above the ground. He throws a large firecracker horizontally with a speed \(v .\) However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude a. This results in the firecracker reaching the ground directly under the student. Determine the height \(h\) in terms of \(v, a,\) and \(g .\) You can ignore the effect of air resistance on the vertical motion.

In a Fourth of July celebration, a firework is launched from ground level with an initial velocity of 25.0 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) from the vertical. At its maximum height it explodes in a starburst into many fragments, two of which travel forward initially at 20.0 \(\mathrm{m} / \mathrm{s}\) at \(\pm 53.0^{\circ}\) with respect to the horizontal, both quantities measured relative to the original firework just before it exploded. With what angles with respect to the horizontal do the two fragments initially move right after the explosion, as measured by a spectator standing on the ground?

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going \(90.0 \mathrm{km} / \mathrm{h},\) to his enemy's car, which is going 110 \(\mathrm{km} / \mathrm{h}\) . The enemy's car is 15.8 \(\mathrm{m}\) in front of the hero's when he lets go of the grenade. If the hero throws the grenade so its initial velocity relative to him is at an angle of \(45^{\circ}\) above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance. Find the magnitude of the velocity both relative to the hero and relative to the earth.

In a World Cup soccer match, Juan is running due north toward the goal with a speed of 8.00 \(\mathrm{m} / \mathrm{s}\) relative to the ground. A teammate passes the ball to him. The ball has a speed of 12.0 \(\mathrm{m} / \mathrm{s}\) and is moving in a direction \(37.0^{\circ}\) east of north, relative to the ground. What are the magnitude and direction of the ball's velocity relative to Juan?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Physics of Motion

Read Explanation

Fluids

Read Explanation

Magnetism

Read Explanation

Measurements

Read Explanation

Translational 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.