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Chapter 3: Problem 52

(II) At what projection angle will the range of a projectile cqual its maximum height?

Short Answer

Expert verified
The projection angle where the range equals the maximum height is approximately 63.4 degrees.

Step by step solution

01

Understanding the Problem

To find the projection angle where the range of a projectile equals its maximum height, we need to use the formula for the range \( R \) and the maximum height \( H \). We know that: \( R = \frac{v^2}{g} \sin(2\theta) \) and \( H = \frac{v^2}{2g} \sin^2(\theta) \), where \( v \) is the initial velocity and \( g \) is the acceleration due to gravity.
02

Setting the Equations Equal

To find the angle where the range equals the maximum height, set \( R = H \). This gives us the equation: \( \frac{v^2}{g} \sin(2\theta) = \frac{v^2}{2g} \sin^2(\theta) \). Simplifying, we get \( 2 \sin(2\theta) = \sin^2(\theta) \).
03

Using Trigonometric Identities

Using the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), rewrite the equation as \( 4\sin(\theta)\cos(\theta) = \sin^2(\theta) \). This further simplifies to \( 4\cot(\theta) = \sin(\theta) \).
04

Solving for \( \theta \)

Divide both sides by \( \cos(\theta) \), this gives \( 4 \cot(\theta) = \tan(\theta) \). Recognizing that \( \cot^2(\theta) = 4 \), we find \( \theta = 63.4^\circ \).

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Key Concepts

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

Projection Angle
In projectile motion, the projection angle is the angle at which an object is launched with a certain speed into the air. It's crucial in determining the path, or trajectory, the object follows. This angle is measured with respect to the horizontal line from the launch point.

The projection angle influences various aspects of the projectile's motion such as its maximum height, range, and time of flight. For instance:
  • A smaller angle results in a lower and longer trajectory.
  • A larger angle leads to a shorter distance covered but a higher peak.
Understanding the projection angle is essential as it helps predict where and how high the projectile will fly.

When solving mathematical problems related to projectiles, it is often necessary to determine at which angle certain conditions, such as the maximum height being equal to the range, are met. This involves using trigonometric identities and algebra to solve for the unknown angle. Such calculations underline the importance of comprehending and manipulating the projection angle to achieve desired results.
Maximum Height
The maximum height of a projectile is the highest vertical position it reaches during its flight. This point occurs when the vertical component of its velocity becomes zero, meaning the projectile momentarily stops rising before it begins to fall back down.

The formula for calculating maximum height ( H ) is given by:
\[ H = \frac{v^2}{2g} \sin^2(\theta) \] where:
  • \( v \) is the initial velocity
  • \( g \) is the acceleration due to gravity
  • \( \theta \) is the projection angle

This equation highlights two key influences on maximum height: the initial speed and the sine of the projection angle. Notice that the height increases with the square of the sine of the angle, making angles close to 90 degrees optimal for achieving higher altitudes.

Understanding how these elements interact is vital for accurately predicting how high a projectile will go, which is particularly important in fields like sports science and engineering.
Range of a Projectile
The range of a projectile refers to the total horizontal distance it covers during its motion from the point of launch to where it lands. Range is a crucial measure, as it determines how far a projectile will travel before hitting the ground.

The formula for the range ( R ) of a projectile is:
\[ R = \frac{v^2}{g} \sin(2\theta) \] where:
  • \( v \) is the initial velocity
  • \( g \) is the acceleration due to gravity
  • \( \theta \) is the projection angle

This equation shows that range depends heavily on both the initial velocity and the sine of twice the projection angle. Interestingly, to maximize the range, a projection angle of 45 degrees is typically ideal when air resistance is negligible.

In this context, finding the specific angle where the range equals the maximum height involves solving complex trigonometric equations. These calculations are critical in real-world applications to ensure that projectiles, whether in ballistics or sports, hit their intended targets efficiently.

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Most popular questions from this chapter

A batter hits a fly ball which leaves the bat \(0.90 \mathrm{~m}\) above the ground at an angle of \(61^{\circ}\) with an initial speed of \(28 \mathrm{~m} / \mathrm{s}\) heading toward centerfield. Ignore air resistance. \((a)\) How far from home plate would the ball land if not caught? \((b)\) The ball is caught by the centerfielder who, starting at a distance of \(105 \mathrm{~m}\) from home plate, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed.

Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height if air resistance is neglible.

In hot pursuit, Agent Logan of the FBI must get directly across a 1200 -m-wide river in minimum time. The river's current is \(0.80 \mathrm{~m} / \mathrm{s}\), he can row a boat at \(1.60 \mathrm{~m} / \mathrm{s}\), and he can run \(3.00 \mathrm{~m} / \mathrm{s}\). Describe the path he should take (rowing plus running along the shore) for the minimum crossing time, and determine the minimum time.

A particle's position as a function of time \(t\) is given by \(\overrightarrow{\mathbf{r}}=\left(5.0 t+6.0 t^{2}\right) \mathrm{m} \hat{\mathbf{i}}+\left(7.0-3.0 t^{3}\right) \mathrm{m} \hat{\mathbf{j}} .\) At \(t=5.0 \mathrm{~s}\) find the magnitude and direction of the particle's displacement vector \(\Delta \overrightarrow{\mathbf{r}}\) relative to the point \(\overrightarrow{\mathbf{r}}_{0}=(0.0 \hat{\mathbf{i}}+7.0 \hat{\mathbf{j}}) \mathrm{m}\)

(III) An airplanc, whose air speed is 580 \(\mathrm{km} / \mathrm{h}\) , is supposed to fly in a straight path \(38.0^{\circ} \mathrm{N}\) of E. But a steady 72 \(\mathrm{km} / \mathrm{h}\) wind is blowing from the north. In what direction should the plane head?
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