/*! 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 21 A man stands on the roof of a 15... [FREE SOLUTION] | 91Ó°ÊÓ

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A man stands on the roof of a 15.0 -m-tall building and throws a rock with a speed of \(30.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(33.0^{\circ}\) above the horizontal. Ignore air resistance. Calculate (a) the maximum height above the roof that the rock reaches; (b) the speed of the rock just before it strikes the ground; and (c) the horizontal range from the base of the building to the point where the rock strikes the ground. (d) Draw \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

Short Answer

Expert verified
a) The maximum height above the roof that the rock reaches is 20.9m. b) The speed of the rock just before it hits the ground is 29.2m/s. c) The horizontal range from the base of the building to the point where the rock strikes the ground is 78.2m. d) The graphs can be drawn as described in step 5.

Step by step solution

01

Split Initial Velocity into Components

First, split the initial velocity of the rock into horizontal \(v_{xi}\) and vertical \(v_{yi}\) components using the angle of projection. Remember, initial horizontal velocity \(v_{xi} = v_i \cdot \cos(\theta)\) and initial vertical velocity \(v_{yi} = v_i \cdot \sin(\theta)\) where \(v_i\) is 30 m/s and \(\theta\) is 33 degrees.
02

Calculate the Maximum Height

Next, calculate the maximum height above the roof that the rock reaches \(h_{max}\). We use this formula for the maximum height in projectile motion: \(h_{max} = (v_{yi}^2) / (2g)\) where \(g\) is acceleration due to gravity (9.81 m/s²). Add the height of the building in the end.
03

Calculate the Speed Just Before Hitting the Ground

For the speed just before it strikes the ground \(v_f\), equate the potential energy at maximum height to kinetic energy just before hitting. Or use the formula derived from kinetic equations of motion: \(v_f = \sqrt{v_{xi}^2 + (v_{yi} + gt)^2}\) where \(t\) is the time of flight that could be found using \(v_{yi} + gt = 0\).
04

Calculate the Horizontal Range

Then, find the horizontal distance or range \(R\) from the base of the building to the point where the rock strikes the ground. We use this formula: \(R = v_{xi} \cdot t\).
05

Draw the Graphs

Finally, sketch the requested \(x-t, y-t, v_{x}-t,\) and \(v_{y}-t\) graphs. The \(x-t\) graph will be a straight line since horizontal velocity is constant in projectile motion without air resistance. The \(y-t\) graph will be a parabola because the vertical movement is accelerated motion. The \(v_{x}-t\) graph is a horizontal straight line (due to constant horizontal velocity), and the \(v_{y}-t\) graph is a downward sloping line (due to constant acceleration from gravity).

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Key Concepts

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

Kinematics
Kinematics refers to the study of motion without considering the forces that cause it. In projectile motion, such as when a rock is thrown from a rooftop, understanding kinematics helps us predict the future position and velocity of the projectile. We focus on breaking down the motion into horizontal and vertical components, as they behave independently.
- The horizontal motion occurs at a constant speed due to no air resistance, making it simple to analyze.
- The vertical motion is influenced by gravity, resulting in a uniformly accelerated motion.
By applying kinematic concepts, we can accurately predict key details like the maximum height, the time of flight, and how far the projectile will travel.
Velocity Components
To understand the motion of the rock more deeply, we need to consider its velocity components. When a projectile is launched at an angle, its initial velocity is divided into two perpendicular components: the horizontal component \(v_{xi}\) and the vertical component \(v_{yi}\).
- **Horizontal Velocity Component**: Found using \(v_{xi} = v_i \cdot \cos(\theta)\). This component remains constant as it moves because no horizontal forces act on the projectile.
- **Vertical Velocity Component**: Calculated using \(v_{yi} = v_i \cdot \sin(\theta)\). This component changes over time due to the gravitational pull, which slows it down until reaching the highest point and then speeds it up as it descends.
Breaking down the velocity in this way simplifies problem-solving by considering each dimension of movement independently.
Projectile Trajectory
The path followed by the projectile is known as its trajectory, which is a curved shape due to the influence of gravity. For our stone, the trajectory is specifically a parabola. This shape occurs because:
- The horizontal motion remains constant and uniform, experiencing no acceleration.
- The vertical motion feels a continuous downward acceleration, due to gravity, creating a symmetrical path.
To calculate the arc's peak, or the maximum height reached above the start point, we use the initial vertical velocity and gravity. The trajectory's horizontal range (distance covered along the ground) is determined by the time the projectile is in motion and its horizontal velocity.
Understanding the trajectory's properties allows us to predict where the rock will land and ensures our calculations consider the influences of gravity and launch angle.
Physics Graphs
Physics graphs visually represent the projectile motion and give insights into different parameters over time. Let's consider the graphs relevant to our problem:
- **Position vs. Time Graphs (x-t and y-t)**: In an x-t graph, which tracks horizontal position over time, you will see a straight line. It means constant horizontal velocity. The y-t graph, showing vertical position, forms a parabola, illustrating acceleration due to gravity.
- **Velocity vs. Time Graphs (\(v_{x}-t\) and \(v_{y}-t\))**: For the \(v_{x}-t\) graph, the velocity is constant, resulting in a horizontal line. For \(v_{y}-t\), it's a straight line with a negative slope, as gravity constantly affects vertical velocity.
By understanding these graphs, we can easily interpret how both position and velocity change over time in projectile motion.

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