/*! 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 46 An egg is thrown nearly vertical... [FREE SOLUTION] | 91Ó°ÊÓ

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An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 30.0 m below its starting point 5.00 \(\mathrm{s}\) after it leaves the thrower's hand. Air resistance may be ignored. (a) What is the initial speed of the egg? (b) How high does it rise above its starting point? (c) What is the magnitude of its velocity at the highest point? (d) What are the magnitude and direction of its acceleration at the highest point? (e) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the egg.

Short Answer

Expert verified
(a) Initial speed is 18.53 m/s. (b) It rises 17.51 m. (c) Velocity at the top is 0 m/s. (d) Acceleration is 9.81 m/s² downwards.

Step by step solution

01

Analyze Given Information

We know the egg passes a point 30.0 m below its initial position 5.00 s after being thrown. This means the egg's displacement, considering it returns below the start point, is \( y = -30.0 \, \text{m} \) and \( t = 5.00 \, \text{s} \). We'll use these details along with equations of motion.
02

Apply Kinematic Equations

We use the equation of motion: \( y = v_i t + \frac{1}{2} a t^2 \). Here, \( y = -30.0 \, \text{m} \), \( t = 5.00 \, \text{s} \), and \( a = -9.81 \, \text{m/s}^2 \) (acceleration due to gravity).Plugging in these values: \( -30.0 = v_i (5.00) - \frac{1}{2} (9.81) (5.00)^2 \). This simplifies to: \( -30.0 = 5.00 v_i - 122.625 \).
03

Solve for Initial Velocity

Re-arranging the equation for \( v_i \): \[ 5.00 v_i = -30.0 + 122.625 \]\[ 5.00 v_i = 92.625 \]\[ v_i = \frac{92.625}{5.00} \approx 18.53 \, \text{m/s} \]Thus, the initial speed of the egg is \( 18.53 \, \text{m/s} \).
04

Determine Maximum Height

At the highest point, velocity \( v = 0 \). Using \( v^2 = v_i^2 + 2a y_{max} \), where \( v = 0 \), \( v_i = 18.53 \, \text{m/s} \), and \( a = -9.81 \, \text{m/s}^2 \), solve for \( y_{max} \):\[ 0 = (18.53)^2 + 2(-9.81) y_{max} \]\[ y_{max} = \frac{(18.53)^2}{2 \times 9.81} \approx 17.51 \, \text{m} \]The egg rises approximately \( 17.51 \, \text{m} \) above the starting point.
05

Calculate Velocity at Highest Point

At the highest point, the velocity of the egg is \( v = 0 \, \text{m/s} \) because it changes direction here.
06

Determine Acceleration at Highest Point

The acceleration due to gravity is constant, so even at the highest point, \( a = 9.81 \, \text{m/s}^2 \) downward.
07

Sketch Graphs

1. **Acceleration vs. Time** (\( a_y - t \)): a constant horizontal line at \(-9.81 \, \text{m/s}^2 \).2. **Velocity vs. Time** (\( v_y - t \)): a straight line decreasing from \( 18.53 \, \text{m/s} \) to \(-30.675 \, \text{m/s} \) at 5 s.3. **Position vs. Time** (\( y - t \)): a parabola opening downwards starting at 0, peaking at \( 17.51 \, \text{m} \), and going to \(-30.0 \, \text{m} \) at 5 s.

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Key Concepts

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

Projectile Motion
Projectile motion is a fascinating concept found in physics, especially when discussing objects launched into the air. It involves analyzing the path of an object like a ball or, in this case, an egg thrown nearly vertically. This motion is affected by two key factors: the initial speed and the force of gravity acting upon the object.
In the scenario given, the egg follows a trajectory that is a result of its upward launch. As it rises, its speed decreases until it reaches the peak of its motion, known as the maximum height. Here, the velocity is zero since it momentarily stops before descending. The overall path resembles a parabola, a curve shaped line as seen in the graph of position versus time in our example.
This motion's intricacies are crucial for calculating related values such as initial velocity, maximum height, and time of flight. Understanding projectile motion is essential for solving problems related to objects moving both vertically and horizontally across different scenarios.
Acceleration Due to Gravity
Gravity is a fundamental force in physics and plays a vital role in determining how objects move near the Earth's surface. The acceleration due to gravity is denoted by the symbol "g," which has a standardized value of approximately \( g = -9.81 \, \text{m/s}^2 \). This negative sign indicates that it is directed downward, toward the center of the Earth.
In the case of the projectile motion of the egg, gravity provides a constant downward acceleration. It affects the egg's motion in two stages: first it decelerates the egg as it rises until it comes to a temporary halt at its highest point, then it accelerates it back to the ground. This consistent force is why the velocity-time graph exhibits a straight line with a negative slope, and why gravity remains essential in calculations such as determining both initial speed and maximum height.
The acceleration due to gravity is constant and does not change regardless of the direction of the motion or altitude, illustrating a critical feature of many kinematic problems.
Velocity-Time Graph
A velocity-time graph provides valuable insights into how an object's speed changes throughout its motion. In the example of the egg, initially launched with a velocity calculated to be approximately \( 18.53 \, \text{m/s} \), the graph starts at this point.
As time progresses, the graph shows a linear decrease because of the constant deceleration provided by gravity. After the initial ascent, the egg’s velocity diminishes to zero, marking the peak of its flight. Beyond this point, the velocity becomes negative, indicating that the egg is now descending.
The slope of this graph represents acceleration; in this scenario, it is a steady negative slope corresponding to the constant acceleration due to gravity \( -9.81 \, \text{m/s}^2 \). By analyzing such graphs, one can easily visualize how velocity is influenced over time by external forces such as gravity in projectile motion situations.
Displacement
Displacement refers to the overall change in an object's position and is a vector quantity. It considers both the distance and the direction between the start and end points of an object's path. In physics, this is crucial for determining the effectiveness of motion and should not be confused with distance, which is scalar and does not account for direction.
From the step-by-step solution, we know that the egg’s displacement is \( -30.0 \, \text{m} \), indicating that it ends up 30 meters lower than its starting point. This negative sign not only hints at the downward direction below its initial launch point but also helps solve related equations for determining initial velocity and time.
Overall, understanding displacement in kinematics enables accurate calculations and predictions regarding an object's motion's final position, which is important when analyzing any sort of projectile motion effectively.

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Most popular questions from this chapter

Large cockroaches can run as fast as 1.50 \(\mathrm{m} / \mathrm{s}\) in short bursts. Suppose you turn on the light in a cheap motel and see one scurrying directly away from you at a constant 1.50 \(\mathrm{m} / \mathrm{s} .\) If you start 0.90 m behind the cockroach with an initial speed of 0.80 \(\mathrm{m} / \mathrm{s}\) toward it, what minimum constant acceleration would you need to catch up with it when it has traveled \(1.20 \mathrm{m},\) just short of safety under a counter?

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