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Chapter 3: Problem 10

Trajectory of particle in a projectile motion is given as: \(y-x-\frac{x^{2}}{80} .\) Here, \(x\) and \(y\) are in metres. For this projectile motion, match the following (with \(g=10 \mathrm{~m} / \mathrm{s}^{2}\). Column-I (a) Angle of projcction (b) Angle of velocity with horizontal after 45 (c) Maximum height (d) Horizontal range Column-II (p) \(20 \mathrm{~m}\) (q) \(80 \mathrm{~m}\) (r) \(45^{\circ}\) (s) \(\tan ^{-1}(1 / 2)\)

Short Answer

Expert verified
a-r; b-s; c-p; d-q.

Step by step solution

01

Understand the Trajectory Equation

The given trajectory equation is \( y = x + \frac{x^2}{80} \). This can be compared to the standard form of a projectile motion equation in terms of \( x \): \( y = x \tan\theta - \frac{g}{2v_0^2\cos^2\theta}x^2 \), where \( \theta \) is the angle of projection and \( v_0 \) is the initial velocity.
02

Determine Angle of Projection

From the equation \( y = x + \frac{x^2}{80} \), comparing with \( y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}x^2 \), the coefficient of \(x\) indicates \( \tan\theta = 1 \), implying \( \theta = 45^{\circ} \). This solves for part (a) to (r).
03

Calculate Maximum Height

The maximum height \(H\) is determined using \( H = \frac{u^2\sin^2\theta}{2g} \). With the trajectory \( x = u^2\sin2\theta/g \), rearranging gives \( u^2 = 80g \), thus \( H = \frac{80g\sin^2\theta}{2g} = 20 \). Thus, part (c) is matched to (p).
04

Calculate Horizontal Range

The horizontal range \(R\) is obtained from \(R = \frac{u^2\sin2\theta}{g}\). Since \( \theta = 45^{\circ} \), \( \sin2\theta = 1\). Therefore, \( R = \frac{80g}{g} = 80 \). Thus, part (d) is matched to (q).
05

Evaluate the Angle After 45 Seconds

To find the angle of velocity with the horizontal after a given time, decompose the velocity vector into horizontal and vertical components. Given \( v_y = u \sin \theta - gt \) after 45 seconds and with \( v_x = u \cos \theta \), \( \theta = 45^{\circ} \), this simplifies to \( \tan^{-1}(1/2) \) for part (b), matching with (s).

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Key Concepts

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

Trajectory Equation
In projectile motion, the trajectory equation tells us the path that an object will take as it moves through the air. For our exercise, the trajectory is described by the equation \( y = x + \frac{x^2}{80} \). This must be compared to the general form \( y = x \tan \theta - \frac{g}{2v_0^2\cos^2\theta}x^2 \).
  • Where: \( y \) is the vertical position, \( x \) is the horizontal position.
  • \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (10 \( \text{m/s}^2 \) in our scenario).
By comparing coefficients, we determine the parameters of the projectile, setting the stage for further calculations. This equation helps us predict where and how high the projectile will go.
Angle of Projection
The angle of projection \( \theta \) is crucial in defining the initial trajectory of a projectile. It determines the direction of the launch relative to the ground. In our problem, the standard trajectory equation is rearranged to reveal \( \tan\theta = 1 \). Consequently, the angle of projection \( \theta \) resolves to \( 45^{\circ} \).
  • Impact: With \( 45^{\circ} \) being a common angle for maximum range in projectiles, it's a key ideal scenario in physics problems.
  • This specific launch angle ensures optimal convexity in travel path, balancing horizontal and vertical components for distance and height.
Understanding the angle helps clarify how far and how high the projectile will travel.
Maximum Height of Projectile
The maximum height is the peak point in the projectile's path. It's found using the formula \( H = \frac{u^2\sin^2\theta}{2g} \). For our exercise, we simplify this to \( H = 20 \) meters.
  • Calculation: We use the relationship \( u^2 = 80 \times g \) based on our trajectory equation, and with \( \theta = 45^{\circ} \), \( \sin\theta = \frac{\sqrt{2}}{2} \), leads us smoothly to the solution.
  • It highlights the movement's vertical component influence at its highest, showing how much kinetic energy went into lifting the projectile.
Knowing the maximum height offers insights into the vertical reach, which could be critical for clearance over obstacles.
Horizontal Range of Projectile
The horizontal range, signified by \( R \), showcases the total distance covered horizontally by the projectile. Calculated by \( R = \frac{u^2 \sin 2\theta}{g} \), it clarifies the extent of the projectile's flight.
  • For this exercise: the horizontal range works out to \( 80 \) meters. Given \( \theta = 45^{\circ} \) leads to maximum \( R \) as \( \sin 2\theta = 1 \), confirming \( R = \frac{80 \times g}{g} = 80 \).
  • The range is maximized at \( 45^{\circ} \), exemplifying the principle that this particular angle allows the largest horizontal expanse.
Understanding the horizontal range helps anticipate where the projectile will land and is essential for planning real-world applications.
Projectile Velocity Components
A projectile's velocity is broken into horizontal and vertical components, explaining how each aspect contributes to motion. At any point in flight, the horizontal velocity \( v_x \) remains constant, whereas the vertical velocity \( v_y \) changes due to gravity.
  • Horizontal velocity: \( v_x = u \cos\theta \).
  • Vertical velocity: \( v_y = u \sin\theta - gt \) (after time \( t \)).
In our problem, after 45 seconds, the vertical velocity adjusts. This is calculated to find the angle of velocity with respect to the horizontal as \( \tan^{-1}(1/2) \). This calculation reveals how gravity progressively tilts the perceived path of motion, crucial in predicting future placement of the projectile.

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Most popular questions from this chapter

A ball is projected horizontally from a height of \(100 \mathrm{~m}\) from the ground with a speed of \(20 \mathrm{~m} / \mathrm{s}\). Find: (a) the time taken to reach the ground, (b) the horizontal distance it covers before striking the ground, and (c) the velocity with which it strikes the ground. Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).

A body is thrown horizontally with a velocity \(\sqrt{2 g h}\) from the top of a tower of height \(h\). It strikes the level ground through the foot of the tower at a distance \(x\) from the tower. Find the value of \(x\).

\(\Lambda\) particle moves along the trajectory of parabola \(y=a x^{2} .\) \Lambdassuming speed to be constant, find the radius of curvature at the origin. (a) \(\frac{1}{2 a}\) (b) \(\frac{1}{a}\) (c) \(\frac{1}{2 a x}\) (d) \(\frac{2}{a}\)

Statement-1: The average velocity of a body in a general projectilc motion during any time interval \(A \leq T\), is uniform, if the body is at the same height above the ground level at two extremities of the time interval \(A t . T\) is the flight time o the projectile Statement-2: 'lhere is no acceleration acting on the projectile in horizontal direction.

\Lambda point on the rim of a circular disc has a linear speed of \(10 \mathrm{~m} / \mathrm{s}\) at an instant when it is decreasing at the rate of \(60 \mathrm{~m} / \mathrm{s}^{2}\). If the magnitude of the total acceleration of the point at this instant is \(100 \mathrm{~m} / \mathrm{s}^{2}\), the radius of the disc is (a) \(1.25 \mathrm{~m}\) (b) \(12.5 \mathrm{~m}\) (c) \(25 \mathrm{~m}\) (d) \(2.5 \mathrm{~m}\)
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