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Chapter 3: Problem 60

Seagulls are often observed dropping clams and other shellfish from a height onto rocks below, as a means of opening the shells. If a seagull drops a shell from rest from a height of \(14 \mathrm{~m}\), how fast is the shell moving when it hits the rocks?

Short Answer

Expert verified
The shell hits the ground with a speed of approximately 16.57 m/s.

Step by step solution

01

Understanding the Problem

We need to determine the final speed of a shell dropped from a height of 14 meters when it hits the ground. The shell starts from rest and is subjected to gravitational force.
02

Identifying Known Values

We have the following known values: the initial velocity of the shell, \(u\), is 0 m/s (since it starts from rest), the height from which it is dropped, \(h\), is 14 m, and the acceleration due to gravity, \(g\), is approximately 9.81 m/s虏.
03

Applying the Kinematic Equation

To find the final velocity, \(v\), we use the kinematic equation for objects in free fall: \[v^2 = u^2 + 2gh\]Substitute the known values into the equation: \[v^2 = 0 + 2(9.81)(14)\]
04

Calculating the Final Velocity

Solving for \(v^2\): \[v^2 = 27.44\times 2\times 14 = 274.68\]Then we take the square root to find \(v\): \[v = \sqrt{274.68} \approx 16.57 \text{ m/s}\]

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Key Concepts

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

Free Fall
When an object is in free fall, it is moving solely under the influence of gravity. No other forces like air resistance interfere with its motion. This means an object in free fall is accelerating due to gravity alone. In our situation, a seagull dropping a shell is a perfect example of free fall. Starting from rest means the shell begins with an initial velocity of 0 m/s. Every second, it speeds up by approximately 9.81 m/s虏, the standard acceleration due to Earth's gravity. Understanding free fall offers insight into how objects move under gravitational influence without other forces altering that motion.
Gravitational Force
Gravitational force plays a central role in controlling the motion of falling objects. It is the attractive force acting between masses and is responsible for pulling objects towards one another. On Earth, gravitational force is the reason why objects fall towards the ground. During the seagull's shell drop, gravity is constantly exerting a downward force on the shell, creating the acceleration we observe. This force is consistent across all free-falling objects near the Earth's surface, making it a fundamental concept in kinematics and physics.
Final Velocity
The final velocity of an object in free fall is the speed it reaches just before impact with the ground. This is calculated using kinematic equations, which help us understand motion in a straightforward way. From a rest position, the shell's initial velocity is 0 m/s. Using the kinematic equation:
  • \[v^2 = u^2 + 2gh\]
we calculate the speed right before the shell hits the rocks. Here, \(u = 0\), \(g = 9.81 \text{ m/s}^2\), and \(h = 14 \text{ m}\). Solving the equation helps us find the final speed with which the shell impacts the ground, providing insights into the dynamics of motion under gravity's influence.

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Most popular questions from this chapter

Predict \& Explain Two bows shoot identical arrows with the same initial speed. To accomplish this, the string in bow 1 must be pulled back farther than the string in bow 2. (a) Is the acceleration of the arrow shot by bow 1 greater than, less than, or equal to the acceleration of the arrow shot by bow 2? (b) Choose the best explanation from among the following: A. The arrow from bow 2 accelerates over a greater distance. B. Both arrows have the same final speed. C. The arrow from bow 1 accelerates over a greater distance.

Two rocks with different weights are dropped on the surface of the Moon (where there is a near vacuum). How does the acceleration of the heavier rock compare to that of the lighter rock?

You throw a ball upward with an initial speed of \(3.0 \mathrm{~m} / \mathrm{s}\) from an initial height of \(1.5 \mathrm{~m}\). After you throw the ball, its acceleration is \(9.81 \mathrm{~m} / \mathrm{s}\) downward. Taking upward to be the positive direction, write the position-time equation for the ball's motion.

Two cars drive on a straight highway. At time \(t=0\), car 1 passes road marker 0 traveling due east with a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). At the same time, car 2 is \(1.0 \mathrm{~km}\) east of road marker 0 traveling at \(30.0 \mathrm{~m} / \mathrm{s}\) due west. Car 1 is speeding up, with an acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\), and car 2 is slowing down, with an acceleration of \(-3.2 \mathrm{~m} / \mathrm{s}^{2}\). (a) Write position-time equations for both cars. Let east be the positive direction. (b) At what time do the two cars meet?

Why is a person doing a cannonball dive an example of free fall while a person descending to Earth on a parachute is not?
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