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Chapter 5: Problem 7

A particle is projected from ground at some angle with the horizontal. Let \(P\) be the point at maximum height \(H\). At what height above the point \(P\) should the particle be aimed to have range equal to maximum height? (1) \(H\) (2) \(2 \mathrm{H}\) (3) \(\mathrm{H} / 2\) (4) \(3 \mathrm{H}\)

Short Answer

Expert verified
Height above P should be \( H \), option (1).

Step by step solution

01

Understanding the Problem

The problem asks us to find the height above point P, the maximum height, where a particle should be aimed such that its range equals this maximum height. This involves using the properties of projectile motion.
02

Using Projectile Motion Formulas

The range of a projectile is given by the formula \( R = \frac{v^2 \sin 2\theta}{g} \) and the maximum height is \( H = \frac{v^2 \sin^2 \theta}{2g} \). To find the specific conditions where the range equals the maximum height, we equate the range, R, to H.
03

Equating Range and Maximum Height

Set the range equal to the height: \( \frac{v^2 \sin 2\theta}{g} = \frac{v^2 \sin^2 \theta}{2g} \). Simplifying this equation will help us find the angle \( \theta \) for which the range equals the height.
04

Solving for the Angle

By equating the range and height as \( \sin 2\theta = \frac{1}{2} \sin^2 \theta \), we substitute \( \sin 2\theta = 2\sin\theta\cos\theta \). This simplifies to solve for \( \theta \). With trigonometric identities, the angle that satisfies this is when \( \theta = 45^{\circ} \).
05

Finding the Target Height Above P

At \( \theta = 45^{\circ} \), the maximum range equals the maximum height. The total maximum height from the initial point, where range equals height, will be twice the maximum height from the ground, meaning the aimed point should be \( H \) above point P, i.e., total height from the ground is \( 2H \). The height above P should be \( H \), which is option (1).

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Key Concepts

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

Maximum Height
Understanding the maximum height in projectile motion provides insights into how high an object can reach when launched at a certain angle and initial velocity. The maximum height, denoted as \(H\), occurs when the vertical component of the projectile's velocity reduces to zero due to gravity. This is the peak of its trajectory.

The formula to find the maximum height is:
  • \( H = \frac{v^2 \sin^2 \theta}{2g} \)
Here:
  • \( v \) is the initial velocity of the projectile,
  • \( \theta \) is the angle of projection, and
  • \( g \) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\).
At this height, the projectile's upward motion is temporarily halted before gravity pulls it back to Earth. Thus, understanding maximum height helps determine the highest point the projectile reaches in its path.
Range of Projectile
The range of a projectile is the total horizontal distance it travels during its flight. This is a key concept for understanding how far a projectile can cover when launched. The range depends on both the initial speed of the projectile and the angle at which it is launched.

The formula for calculating the range is:
  • \( R = \frac{v^2 \sin 2\theta}{g} \)
In this equation:
  • \( v \) is the initial velocity of the projectile,
  • \( \theta \) is the angle of projection, and
  • \( g \) is the acceleration due to gravity.
Importantly, the angle \( \theta \) significantly affects the range. For example, a 45-degree angle of projection typically yields the maximum range for a given initial velocity. This is because the horizontal and vertical components of the projectile's initial velocity are equal at this angle, balancing the dynamics optimally. Understanding the range helps predict the overall path and landing point of the projectile.
Angle of Projection
The angle of projection, \( \theta \), is the angle at which a projectile is launched relative to the horizontal plane. This angle has a significant impact on both the maximum height and the range of the projectile.

Different angles yield different trajectories:
  • Small angles close to 0掳 produce a low, long trajectory.
  • Large angles close to 90掳 produce a high, short trajectory.
  • An angle of 45掳 generally offers the maximum range with balanced height.
When solving problems involving projectile motion, understanding how to manipulate the angle of projection is crucial to meeting specific requirements, such as achieving a particular range. In the exercise we discussed, we solved for the angle \( \theta = 45^{\circ} \) at which the range equaled the maximum height. This demonstrates how critical the angle of projection is in determining a projectile's flight path.

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Most popular questions from this chapter

A river is flowing towards with a velocity of \(5 \mathrm{~m} \mathrm{~s}^{-1}\). The boat velocity is \(10 \mathrm{~m} \mathrm{~s}^{-1}\). The boat crosses the river by shortest path. Hence, (1) The direction of boat's velocity is \(30^{\circ}\) west of north. (2) The direction of boat's velocity is north-west. (3) Resultant velocity is \(5 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\). (4) Resultant velocity of boat is \(5 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\).

Ram and Shyam are walking on two perpendicular tracks with speed \(3 \mathrm{~ms}^{-1}\) and \(4 \mathrm{~m} \mathrm{~s}^{-1}\), respectively. At a certain moment (say \(\mathrm{t}=0 \mathrm{~s}\) ), Ram and Shyam are at \(20 \mathrm{~m}\) and \(40 \mathrm{~m}\) away from the intersection of tracks, respectively, and moving towards the intersection of the tracks. The time \(t\) when they'are at shortest distance from each other subsequently, is - (1) \(8.8 \mathrm{~s}\) (2) \(12 \mathrm{~s}\) (3) \(15 \mathrm{~s}\) (4) \(44 \mathrm{~s}\)

A body is thrown with the velocity \(v_{0}\) at an angle of \(\theta\) to the horizon. Determine \(v_{0}\) in \(\mathrm{ms}^{-1}\) if the maximum height attained by the body is \(5 \mathrm{~m}\) and at the highest point of its trajectory the radius of curvature is \(r=3 \mathrm{~m}\). Neglect air resistance. [Use \(\sqrt{80}\) as 9 ]

A train of \(150 \mathrm{~m}\) length is going toward north direction at a speed of \(10 \mathrm{~m} \mathrm{~s}^{-1}\). A parrot flies at a speed of \(5 \mathrm{~m} \mathrm{~s}^{-1}\) toward south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to (1) \(12 \mathrm{~s}\) (2) \(8 \mathrm{~s}\) (3) \(15 \mathrm{~s}\) (4) \(10 \mathrm{~s}\)

A particle is moving in \(x-y\) plane. At certain instant of time, the components of its velocity and acceleration are as follows: \(v_{x}=3 \mathrm{~m} / \mathrm{s}, v_{y}=4 \mathrm{~m} / \mathrm{s}, a_{\mathrm{x}}=2 \mathrm{~m} / \mathrm{s}^{2}\) and \(a_{\mathrm{y}}=1 \mathrm{~m} / \mathrm{s}^{2} .\) Find the rate of change of speed (in \(\mathrm{m} / \mathrm{s}^{2}\) ) at this moment.
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