/*! 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 25 The best leaper in the animal ki... [FREE SOLUTION] | 91Ó°ÊÓ

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The best leaper in the animal kingdom is the puma, which can jump to a height of \(12 \mathrm{ft}\) when leaving the ground at an angle of \(45^{\circ}\). With what speed, in SI units, must the animal leave the ground to reach that height?

Short Answer

Expert verified
The puma must leave the ground with a speed of approximately \(23.34 \, m/s\)

Step by step solution

01

Convert ft to m

First we need to convert the given height from feet to meters. We know that 1 ft = 0.3048 m. So, \(12 \, ft = 12 \times 0.3048 = 3.66 \, m\)
02

Convert Degrees to Radians

Next convert the given angle from degrees to radians. As there are \(\pi\) radians in \(180 \, degrees\), we have \(45 \, degrees = 45 \times \frac{\pi}{180} = \frac{\pi}{4} \, radians\)
03

Solve for Speed

Now we can substitute these values into the equation for \(h\) to solve for \(v\). Rearranging the equation gives \(v = \sqrt{2gh/(sin^2(\theta))} = \sqrt{2 \times 9.8 \times 3.66 / \sin^2(\pi/4)} = \sqrt{2 \times 9.8 \times 3.66 / 0.5} \approx 23.34 \, m/s\)
04

Cross Check

The result is cross-checked for validity by substituting it back into the original equation to ensure the correct height is reached.

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Key Concepts

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

Conversion of Units
Often when solving physics problems, you'll need to convert units to ensure consistency. In projectile motion problems, it's common to encounter measurements in various unit systems. Here's how you can manage these conversions effectively.

If the initial values are in feet, like in the exercise with the puma's jump, it's necessary to convert these into meters, especially since the SI unit system uses meters for length. Feet can be converted to meters using the factor:
  • 1 foot = 0.3048 meters
Thus, to convert the 12 feet to meters, multiply by the conversion factor:
  • \[12 \times 0.3048 = 3.66 \text{ meters}\]
This allows the calculations to remain consistent with the SI units, providing accurate results.

Similarly, if angles are given in degrees, converting to radians is crucial for trigonometric calculations. Degrees can be converted to radians as:
  • Radian measure = Degree measure × \( \frac{\pi}{180} \)
This makes it easier to perform calculations with trigonometric functions, which inherently use radians in their mathematical definitions.
Trigonometry in Physics
Trigonometry is pivotal in analyzing projectile motion, as it helps break down motion into horizontal and vertical components. This is especially useful when an object is launched at an angle.

When a projectile, like a puma leaping at an angle of 45 degrees, you need trigonometric functions to resolve the initial velocity into vertical and horizontal components. The key functions used are:
  • Sine (sin) for vertical calculations
  • Cosine (cos) for horizontal calculations
For an angle \( \theta \) with an initial speed \( v \):

  • Vertical component \( v_y = v \cdot \sin(\theta) \)
  • Horizontal component \( v_x = v \cdot \cos(\theta) \)
In a scenario like the puma's jump, where the angle is a standard case of 45 degrees, both the sine and cosine of 45 degrees will equal 1/\( \sqrt{2} \), simplifying the calculations and making the analysis straightforward.

Understanding trigonometry's role in physics allows you to efficiently decompose and analyze motion, providing clarity and insight into the problem.
Kinematics
Kinematics focuses on the motion of objects without considering the forces that cause this motion. In problems involving projectiles, understanding kinematic equations can vastly simplify solving for unknown values.

In our specific problem, the main kinematic equation used relates the maximum height \( h \), the initial velocity \( v \), and the angle \( \theta \): \[ h = \frac{v^2 \cdot \sin^2(\theta)}{2g} \] where:
  • \( h \) is the maximum height
  • \( v \) is the initial velocity
  • \( g \) is the acceleration due to gravity (approximated as \( 9.8 \text{ m/s}^2 \))
By rearranging this equation, we can solve for the initial velocity \( v \): \[ v = \sqrt{\frac{2gh}{\sin^2(\theta)}} \] For the puma, substituting in the values, we find that:\[ v = \sqrt{\frac{2 \times 9.8 \times 3.66}{0.5}} \approx 23.34 \text{ m/s} \]

Mastering kinematics means having a firm grasp on how objects move, giving you the tools to analyze any trajectory, whether in simple school exercises or complex real-world scenarios.

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Most popular questions from this chapter

From the window of a building, a ball is tossed from a height \(y_{0}\) above the ground with an initial velocity of \(8.00 \mathrm{~m} / \mathrm{s}\) and angle of \(20.0^{\circ}\) below the horizontal. It strikes the ground \(3,00 \mathrm{~s}\) later, \((\mathrm{a})\) If the base of the building is taken to be the origin of the coordinates, with upward the positive \(y\) -direction, what are the initial coordinates of the ball? (b) With the positive \(x\) -direction chosen to be out the window, find the \(x\) -and \(y\) -components of the initial velocity. (c) Find the equations for the \(x\) -and \(y\) components of the position as functions of time. (d) How far horizontally from the base of the building does the ball strike the ground? (e) Find the height from which the ball was thrown. (f) How long does it take the ball to reach a point \(10.0 \mathrm{~m}\) below the level of launching?

One of the fastest recorded pitches in major-league baseball, thrown by Joel Zumaya in 2006, was clocked at \(101.0 \mathrm{mi} / \mathrm{h}\) (Fig. P3.22). If a pitch were thrown horizonLally with this velocity, how far would the ball fall vertically by the time it reached home plate, \(60.5 \mathrm{ft}\) away?

You can use any coordinate system you like to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity \(\vec{v}\) at an angle \(\theta\) with respect to the horizontal. Let the building be \(50.0 \mathrm{~m}\) tall, the initial horizontal velocity be \(9.00 \mathrm{~m} / \mathrm{s}\), and the initial vertical velocity be \(12.0 \mathrm{~m} / \mathrm{s}\). Choose your coordinates such that the positive \(y\) -axis is upward, the \(x\) -axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball's maximum height above the ground and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.

A girl delivering newspapers covers her route by traveling \(3.00\) blocks west, \(4.00\) blocks north, and then \(6.00\) blocks east. (a) What is her resultant displacement? (b) What is the total distance she travels?

A dart gun is fired while being held horizontally at a height of \(1.00 \mathrm{~m}\) above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance of \(5.00 \mathrm{~m}\). A college student holds the same gun in a horizontal position while sliding down a \(45.0^{\circ}\) incline at a constant speed of \(2.00 \mathrm{~m} / \mathrm{s}\). How far will the dart travel if the student fires the gun when it is \(1.00 \mathrm{~m}\) above the ground?

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