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Chapter 3: Problem 64

Henrietta is jogging on the sidewalk at \(3.05 \mathrm{~m} / \mathrm{s}\) on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is \(38.0 \mathrm{~m}\) above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally \(9.00 \mathrm{~s}\) after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

Short Answer

Expert verified
(a) The bag must be thrown with an initial speed of approximately 3.05 m/s. (b) Henrietta is approximately 35.96 meters away from her starting point when she catches the bag.

Step by step solution

01

Determine the Time it Takes for the Bag to Reach the Ground

Since the bag is thrown from a height of 38.0 m and subject only to the acceleration due to gravity (since we are ignoring air resistance), we can use the second equation of motion: \(h = ut + \frac{1}{2}gt^2\) where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time. Since the bag is thrown horizontally, the initial vertical velocity (u) is 0. Hence, the equation simplifies to \(38 = \frac{1}{2}(9.8)t^2\). Solving for t gives us t = \( \sqrt{\frac{38 \times 2}{9.8}} \approx 2.79 \text{ seconds}\).
02

Determine the Initial Speed of the Bag

In this vertical motion problem, the bag must be thrown such that it reaches Henrietta in 2.79 seconds. Hence, the horizontal distance covered by the bag in this time is the same as the distance covered by Henrietta in 2.79 seconds. Since Henrietta's speed is 3.05 m/s, she would cover a distance of \(3.05 \times 2.79 \approx 8.51 \text{ meters}\) in 2.79 seconds. Therefore, the initial speed of the bag must be such that it also covers this distance in the same time. So the initial speed (v) of the bag is \( \frac{s}{t} = \frac{8.51}{2.79} \approx 3.05 \text{ m/s}\).
03

Determine Henrietta's Position

Here, we must establish where Henrietta is when she catches the bag. She started at the point directly below the window and has been running at 3.05 m/s. When the bag is thrown, it has already been 9.00 seconds since Henrietta started jogging. Thus, she will be \(3.05 \times 9 = 27.45 \text{ meters}\) away from the starting position at the time the bag is thrown. Since, as determined above, the bag takes 2.79 seconds to reach Henrietta, during this time Henrietta covers an additional \(3.05 \times 2.79 \approx 8.51 \text{ meters}\). Therefore, when she catches the bag, Henrietta is \(27.45 + 8.51 = 35.96 \text{ meters}\) from the starting position.

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Key Concepts

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

Projectile Motion
Projectile motion describes the motion of an object that is thrown, or projected, into the air and is subject only to the force of gravity. It's a combination of two simultaneous yet independent motions: the horizontal motion, with no acceleration (assuming no air resistance), moves at a constant speed; and the vertical motion, which is subject to the constant acceleration due to gravity.

In Henrietta's example, when Bruce throws the bag of bagels, it exhibits projectile motion. It moves horizontally with the initial speed Bruce gives it and at the same time, it accelerates down towards Earth due to gravity. The key to solving the exercise is to treat the two motions separately, then combine them to find where Henrietta will be when she catches the bag.
Equations of Motion
The equations of motion are a set of formulas that describe the relationship between displacement, velocity, acceleration, and time. They are particularly useful when analyzing objects moving with uniform acceleration 鈥� like the acceleration due to gravity in projectile motion.

In the context of the bag thrown to Henrietta, we use the second equation of motion to determine the time it takes for the bag to reach her, which does not consider the horizontal component of its motion. This simplifies the problem to a one-dimensional fall under gravity, letting us find the time of flight exclusively from the vertical motion.
Acceleration Due to Gravity
On Earth, all objects accelerate towards the ground at the same rate (ignoring air resistance) due to gravity. This acceleration is approximately 9.8 meters per second squared (m/s虏). Importantly, this affects only the vertical component of an object's motion, not the horizontal.

For Bruce's bag of bagels, we consider the acceleration due to gravity when calculating the time for the bag to hit the ground. Assuming initial vertical velocity is zero because the bag is thrown horizontally, the bag鈥檚 vertical descent is solely due to Earth鈥檚 gravitational pull.
Initial Velocity
The initial velocity of an object is the velocity at the start of a time period under consideration. In two-dimensional motion, like projectile motion, initial velocity has both a horizontal and a vertical component. However, if there is only horizontal motion at the start, such as in the bagel-tossing problem, the initial vertical velocity is zero.

The horizontally thrown bag's initial velocity must be calculated to ensure it reaches Henrietta before it hits the ground. This factor is critical, as it determines how far the bag will travel horizontally while it falls. The horizontal distance covered by the bag, which in this case should equal the distance Henrietta travels while jogging, helps us determine the bag's necessary initial velocity.

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Most popular questions from this chapter

In Canadian 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, \(36.0 \mathrm{ft}\) horizontally from the bar (Fig. \(\mathbf{P 3 . 6 0}\) ). Football regulations are stated in English units, but convert them 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 is just to clear the bar? Express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h}\).

A roller coaster car moves in a vertical circle of radius \(R\). At the top of the circle the car has speed \(v_{1}\), and at the bottom of the circle it has speed \(v_{2},\) where \(v_{2}>v_{1} .\) (a) When the car is at the top of its circular path, what is the direction of its radial acceleration, \(a_{\mathrm{rad}, \text { top }} ?\) (b) When the car is at the bottom of its circular path, what is the direction of its radial acceleration, \(a_{\mathrm{rad}, \text { bottom }} ?\) (c) In terms of \(v_{1}\) and \(v_{2}\), what is the ratio \(a_{\text {rad, bottom }} / a_{\text {rad, top }} ?\)

The earth has a radius of \(6380 \mathrm{~km}\) and turns around once on its axis in \(24 \mathrm{~h}\). (a) What is the radial acceleration of an object at the earth's equator? Give your answer in \(\mathrm{m} / \mathrm{s}^{2}\) and as a fraction of \(g .\) (b) If \(a_{\mathrm{rad}}\) at the equator is greater than \(g\), objects will fly off the earth's surface and into space. (We'll see the reason for this in Chapter 5.) What would the period of the earth's rotation have to be for this to occur?

A squirrel has \(x\) - and \(y\) -coordinates \((1.1 \mathrm{~m}, 3.4 \mathrm{~m})\) at time \(t_{1}=0\) and coordinates \((5.3 \mathrm{~m},-0.5 \mathrm{~m})\) at time \(t_{2}=3.0 \mathrm{~s}\). For this time interval, find (a) the components of the average velocity, and (b) the magnitude and direction of the average velocity.

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in \(2.70 \mathrm{~s},\) while Milada jumps horizontally with an initial speed of \(95.0 \mathrm{~cm} / \mathrm{s}\). How far from the base of the cliff will Milada hit the ground? Ignore air resistance.
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