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Chapter 3: Problem 15

A skateboarder shoots off a ramp with a velocity of \(6.6 \mathrm{m} / \mathrm{s}\), directed at an angle of \(58^{\circ}\) above the horizontal. The end of the ramp is \(1.2 \mathrm{m}\) above the ground. Let the \(x\) axis be parallel to the ground, the \(+y\) direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

Short Answer

Expert verified
(a) Highest point above the ground is 2.79 m. (b) Horizontal distance is 1.99 m.

Step by step solution

01

Decompose the Initial Velocity

The initial velocity of the skateboarder is given as \(6.6\, \text{m/s}\) at an angle of \(58^\circ\) above the horizontal. We need to find its components.- The horizontal component (\(v_{0x}\)) is given by \(v_{0x} = 6.6 \times \cos(58^\circ)\).- The vertical component (\(v_{0y}\)) is given by \(v_{0y} = 6.6 \times \sin(58^\circ)\).Let's calculate these components:\[ v_{0x} = 6.6 \times \cos(58^\circ) \approx 3.49 \text{ m/s} \]\[ v_{0y} = 6.6 \times \sin(58^\circ) \approx 5.58 \text{ m/s} \]
02

Calculate the Maximum Height Above the Ramp

To find the maximum height above the ramp, we'll use the formula for the vertical motion:\[ v_{y}^2 = v_{0y}^2 - 2g(h - h_0) \]At the highest point, the final vertical velocity (\(v_{y}\)) is 0. Thus,\[ 0 = v_{0y}^2 - 2g(h - h_0) \]Rearranging gives,\[ h - h_0 = \frac{v_{0y}^2}{2g} \]Calculating this:\[ h - h_0 = \frac{(5.58)^2}{2 \times 9.81} \approx 1.59 \text{ m} \]This is the height reached above the 1.2 m, so the maximum height above the ground is:\[ h = 1.2 + 1.59 = 2.79 \text { m} \]
03

Calculate Horizontal Distance at Maximum Height

We need to find the time taken to reach the maximum height first. We know:\[ v_y = v_{0y} - gt \]Setting \(v_y = 0\) to find the time to the highest point (\(t_{max}\)):\[ 0 = 5.58 - 9.81t \]\[ t_{max} = \frac{5.58}{9.81} \approx 0.57 \text{s} \]Now, calculate the horizontal distance (\(x\)) traveled during this time:\[ x = v_{0x} \times t_{max} \]\[ x = 3.49 \times 0.57 \approx 1.99 \text{ m} \]
04

Finalize the Answers

For Part (a), the highest point the skateboarder reaches above the ground is \(2.79\, \text{m}\).For Part (b), the horizontal distance from the end of the ramp at this highest point is \(1.99\, \text{m}\).

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces that cause it. In our skateboarder scenario, we use kinematic equations to determine different aspects of the projectile's path. These equations help us decompose and analyze the skateboarder's motion into horizontal and vertical components.
When an object moves at an angle, like the skateboarder, it's essential to break down its velocity into two parts:
  • Horizontal component (\(v_{0x}\)): which is the velocity in the direction parallel to the ground.
  • Vertical component (\(v_{0y}\)): which is the velocity in the direction upwards, perpendicular to the ground.
Decomposing the velocity is done using trigonometric functions such as sine and cosine, which relate the angle to the components.
These components allow us to apply kinematic equations separately to the horizontal and vertical motions. This separation is crucial because, in projectile motion, each component behaves independently: the horizontal motion has a constant velocity, while the vertical motion is influenced by gravity.
Trajectory Calculations
Trajectory calculations involve predicting and analyzing the path taken by a projectile. In this problem, we're interested in finding the maximum height and horizontal distance the skateboarder travels.
To calculate these, first, determine the maximum height reached by using the kinematic equation for vertical motion, that considers the initial vertical velocity (\(v_{0y}\)) and the effects of gravity. At the peak, where the skateboarder reaches the maximum height, the vertical velocity is zero.
Using the equation \[ v_{y}^2 = v_{0y}^2 - 2g(h - h_0) \] at the highest point lets us calculate the maximum height above the ramp. We add this height to the height of the ramp itself to find the skateboarder's maximum height above the ground.
  • This total height accounts for both the initial elevation of the ramp and the extra height gained during the jump.
The horizontal distance at this point is calculated by determining how long it took the skateboarder to reach the apex of their trajectory and then finding how far they've traveled in that time using the horizontal component of their velocity. This time is determined by setting the vertical velocity to zero and solving for time.
Physics Problem Solving
Physics problem solving is about breaking down complex scenarios into understandable and manageable parts. This allows us to use basic physics principles to find solutions.
In projectile motion problems, like with the skateboarder, it's vital to:
  • Identify what is given (initial velocity, angle, and elevation).
  • Determine what to find (maximum height and horizontal distance).
  • Choose appropriate physics equations.
Doing so usually involves applying kinematics principles to both vertical and horizontal motion components separately. For the vertical, the effects of gravity are pivotal, while the horizontal component relies on the assumption of constant velocity.
By solving each section systematically, checking calculations, and understanding the relationships between different parts of motion, we can effectively solve these problems. In addition, being meticulous about units and proper use of trigonometry can help avoid errors. Practicing these methods enhances problem-solving skills and conceptual understanding, making physics less intimidating and more engaging.

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Most popular questions from this chapter

A golfer imparts a speed of \(30.3 \mathrm{m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the green?

A spacecraft is traveling with a velocity of \(v_{0 x}=5480 \mathrm{m} / \mathrm{s}\) along the \(+x\) direction. Two engines are turned on for a time of 842 s. One engine gives the spacecraft an acceleration in the \(+x\) direction of \(a_{x}=1.20 \mathrm{m} / \mathrm{s}^{2}\), while the other gives it an acceleration in the \(+y\) direction of \(a_{y}=8.40 \mathrm{m} / \mathrm{s}^{2}\) At the end of the firing, find (a) \(v_{x}\) and (b) \(v_{y^{*}}\)

In a stunt being filmed for a movie, a sports car overtakes a truck towing a ramp, drives up and off the ramp, soars into the air, and then lands on top of a flat trailer being towed by a second truck. The tops of the ramp and the flat trailer are the same height above the road, and the ramp is inclined \(16^{\circ}\) above the horizontal. Both trucks are driving at a constant speed of \(11 \mathrm{m} / \mathrm{s}\), and the flat trailer is \(15 \mathrm{m}\) from the end of the ramp. Neglect air resistance, and assume that the ramp changes the direction, but not the magnitude, of the car's initial velocity. What is the minimum speed the car must have, relative to the road, as it starts up the ramp?

Multiple-Concept Example 4 deals with a situation similar to that presented here. A marble is thrown horizontally with a speed of \(15 \mathrm{m} / \mathrm{s}\) from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of \(65^{\circ}\) with the horizontal. From what height above the ground was the marble thrown?

Two cannons are mounted as shown in the drawing and rigged to fire simultaneously. They are used in a circus act in which two clowns serve as human cannonballs. The clowns are fired toward each other and collide at a height of \(1.00 \mathrm{m}\) above the muzzles of the cannons. Clown \(\mathrm{A}\) is launched at a \(75.0^{\circ}\) angle, with a speed of \(9.00 \mathrm{m} / \mathrm{s}\). The horizontal separation between the clowns as they leave the cannons is \(6.00 \mathrm{m}\). Find the launch speed \(v_{0 \mathrm{B}}\) and the launch angle \(\theta_{\mathrm{B}}\left(>45.0^{\circ}\right)\) for clown \(\mathrm{B} .\)
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