/*! 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 49 The angle of projection of a pro... [FREE SOLUTION] | 91Ó°ÊÓ

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The angle of projection of a projectile for which the horizontal range and maximum height are equal to (a) \(\tan ^{-1}(2)\) (b) \(\tan ^{-1}(4)\) (c) \(\cot ^{-1}(2)\) (d) \(60^{\circ}\)

Short Answer

Expert verified
The correct angle is \( \cot^{-1}(2) \), option (c).

Step by step solution

01

Understanding the Problem

We need to determine the angle of projection for which the horizontal range and the maximum height of a projectile are equal. We have four multiple-choice answers to choose from.
02

Determine the Equations of Range and Height

The horizontal range \( R \) of a projectile is given by \( R = \frac{u^2 \sin(2\theta)}{g} \), and the maximum height \( H \) is given by \( H = \frac{u^2 \sin^2(\theta)}{2g} \) where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.
03

Set Range Equal to Height

Since the problem states that the horizontal range and the maximum height are equal, we equate the two expressions: \( \frac{u^2 \sin(2\theta)}{g} = \frac{u^2 \sin^2(\theta)}{2g} \).
04

Simplify the Equation

Cancel \( u^2/g \) from both sides to get \( \sin(2\theta) = \frac{1}{2}\sin^2(\theta) \). Using the identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), substitute to form \( 2\sin(\theta)\cos(\theta) = \frac{1}{2}\sin^2(\theta) \).
05

Divide and Rearrange

Divide both sides by \( \sin(\theta) \) (assuming \( \sin(\theta) eq 0 \)) to obtain \( 2\cos(\theta) = \frac{1}{2}\sin(\theta) \). Rearrange to get \( 4\cos(\theta) = \sin(\theta) \).
06

Solve for \( \theta \)

Taking the tangent of both sides gives \( \tan(\theta) = \frac{1}{4} \), leading to \( \theta = \tan^{-1}\left(\frac{1}{4}\right) \). However, this is not among the given options. Consider complementary angles \( \theta = \cot^{-1}(2) \).
07

Final Answer

The correct angle of projection is \( \theta = \cot^{-1}(2) \), which corresponds to option (c).

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Key Concepts

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

Angle of Projection
When a projectile is launched, the angle at which it is projected into the air is called the **angle of projection**. This angle, denoted by \(\theta\), has a significant impact on how the projectile will travel. It's typically measured from the horizontal line where the projectile begins its journey. Understanding the angle of projection is crucial for predicting the trajectory and final landing point of the projectile.

In the context of projectile motion, your choice of the angle can determine key outcomes, like maximizing distance or height attained. For instance, launching at a 45° angle sans air resistance yields the greatest horizontal range. However, when variables like air resistance or specific requirements such as equal height and distance are in play, the optimal angle can vary.
  • The angle helps dictate the relationship between horizontal and vertical components of velocity.
  • Different angles will affect the symmetry and duration of the projectile’s flight.
  • It influences both the horizontal range and maximum height, key calculations in projectile motion problems.
Horizontal Range
**Horizontal range** is a crucial component of projectile motion. It refers to the total distance a projectile covers over the ground from its initial launch point until it lands. For projectiles launched on level ground under ideal conditions—meaning there’s no air resistance and the landing height equals the launch height—this can be determined by the equation:

\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Here, \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.
  • The range depends strongly on both the initial velocity and the angle of projection.
  • Adjusting \(\theta\) alters \(\sin(2\theta)\), which in turn influences how far the projectile will travel.
  • The range is maximized at a 45° angle, which is the perfect balance between upward and forward momentum.
In cases where needs dictate equal horizontal and vertical travel, certain calculations, as seen in the original exercise, become necessary to identify the right \(\theta\).
Maximum Height
The **maximum height** of a projectile is another important factor to consider. It represents how high the projectile will climb into the air before gravity pulls it back down. This height is determined by:\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]Here, \( u \) is again the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

Understanding the maximum height can help in designing factors like barriers or predicting when and where the projectile might face obstacles.
  • As in the context of the original exercise, when a projectile’s horizontal range equals its maximum height, specific calculations can determine \(\theta\).
  • The maximum height increases with higher angles up to 90°, although practical considerations often prevent such strategies.
  • Factors such as initial speed and projection angle closely influence maximum height—crucial in optimizing flight for different scenarios.
Accurate calculations of maximum height can aid in an array of practical applications, from sports to engineering.
Trigonometric Identities
**Trigonometric identities** play an essential role in solving projectile motion problems. They provide the mathematical tools to manipulate and simplify the equations involved in projectile motion. Key identities often used in these calculations include:

\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \] and the relationship \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).
These identities help break down complex projectile motion equations into simpler, solvable forms.
  • They allow for the conversion between different trigonometric functions depending on what's needed in your calculations.
  • They are instrumental in deriving relationships, as in the given problem where equating range and height required utilizing identities.
  • Understanding these identities can turn a seemingly difficult problem into something manageable, fostering greater insight into the motion of projectiles.
With practice, using trigonometric identities becomes intuitive, especially on tasks that require funnelling projectile constraints into easily interpretable solutions.

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Most popular questions from this chapter

Two particles are projected in air with speed \(v_{0}\) at angles \(\theta_{1}\) and \(\theta_{2}\) (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (a) angle of projection : \(\theta_{1}>\theta_{2}\) (b) time of flight : \(T_{1}>T_{2}\) (c) horizontal range : \(R_{1}>R_{2}\) (d) total energy : \(U_{1}>U_{2}\)

A bomber plane moves horizontally with a speed of \(500 \mathrm{~ms}^{-1}\) and a bomb releases from it, strikes the ground in 10 s. Angle at which it strikes the ground will be \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(\tan ^{-1}\left(\frac{1}{5}\right)\) (b) \(\tan \left(\frac{1}{5}\right)\) (c) \(\tan ^{-1}(1)\) (d) \(\tan ^{-1}(5)\)

A projectile is launched with a speed of \(10 \mathrm{~m} / \mathrm{s}\) at an angle \(60^{\circ}\) with the horizontal from a sloping surface of inclination \(30^{\circ} .\) The range \(R\) is. (Take, \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(4.9 \mathrm{~m}\) (b) \(13.3 \mathrm{~m}\) (c) \(9.1 \mathrm{~m}\) (d) \(12.6 \mathrm{~m}\)

The ceiling of a long hall is \(25 \mathrm{~m}\) high. Then, the maximum horizontal distance that a ball thrown with a speed of \(40 \mathrm{~m} / \mathrm{s}\) can go without hitting the ceiling of the hall, is [NCERT Exemplar] (a) \(95.5 \mathrm{~m}\) (b) \(105.5 \mathrm{~m}\) (c) \(100 \mathrm{~m}\) (d) \(150.5 \mathrm{~m}\)

Assertion-Reason type. Each of these contains two Statements: Statement I (Assertion), Statement II (Reason). Each of these questions also has four alternative choices, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given ahead (a) If both Assertion and Reason are true and Reason is correct explanation of the Assertion (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion (c) If Assertion is true but Reason is false (d) If Assertion is false but the Reason is true Assertion If a particle is projected vertices upwards with velocity \(u\), the maximum height attained by the particle is \(h_{1}\). The same particle is projected at angle \(30^{\circ}\) from horizontal with the same speed \(u\). Now the maximum height is \(h_{2}\). Thus, \(h_{1}=4 h_{2}\) Reason In first case \(v=0\) at highest point and in second case \(v \neq 0\) at highest point.

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