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Chapter 10: Problem 10

A particle is projected with initial speed \(u\) at an angle \(\alpha\) to the horizontal. Its horizontal range is \(R\). (a) The particle would have the same horizontal range if the angle of projection was \(90^{\circ}-\alpha\) (b) The time of flight is \(\frac{R}{u \cos \alpha}\) (c) The particle reaches a maximum height of \(\frac{u^{2} \sin ^{2} \alpha}{2 g}\) above its point of projection.

Short Answer

Expert verified
All three statements (a), (b), and (c) regarding the projectile motion are correct.

Step by step solution

01

Understand the Given Problem

A particle is projected with initial speed \(u\) at an angle \(\alpha\) to the horizontal. The horizontal range is given as \(R\). The task is to verify three statements regarding the projectile motion.
02

Verify Statement (a)

Statement (a) claims that the particle would have the same horizontal range if the angle of projection was \(90^{\circ} - \alpha\). Using the range formula: \[ R = \frac{u^2 \sin(2\alpha)}{g} \] We need to check if: \[ R = \frac{u^2 \sin(2(90^{\circ} - \alpha))}{g} \] realized that \(\sin(2(90^{\circ} - \alpha)) = \sin(180^{\circ} - 2\alpha) = \sin(2\alpha)\), proving the ranges are equal.
03

Verify Statement (b)

Statement (b) claims the time of flight is \(\frac{R}{u \cos \alpha}\). The time of flight formula is: \[ T = \frac{2u \sin \alpha}{g} \] Since: \[ R = \frac{u^2 \sin(2\alpha)}{g} = \frac{u^2 \cdot 2 \sin\alpha \cos\alpha}{g} = \frac{2u^2 \sin\alpha \cos\alpha}{g} \] Rearranging, \[ R = \frac{u \cos\alpha \cdot \frac{2u \sin \alpha}{g}} = u \cos \alpha \cdot T \Rightarrow T = \frac{R}{u \cos \alpha} \]
04

Verify Statement (c)

Statement (c) claims the maximum height is \(\frac{u^2 \sin ^2 \alpha}{2 g}\). Using the maximum height formula: \[ H = \frac{u^2 \sin^2\alpha}{2g} \] The given statement matches the derived formula, therefore it is true.

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Key Concepts

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

Initial Speed
The initial speed, often denoted as \(u\), is the speed with which a particle is projected into the air. This speed splits into two components: horizontal and vertical. For a particle projected at an angle \(\alpha\):
  • Horizontal component: \(u \cos\alpha\)
  • Vertical component: \(u \sin\alpha\)
The initial speed strongly influences the projectile's range, time of flight, and maximum height. Understanding these components helps predict the particle's trajectory.
Angle of Projection
The angle of projection, represented by \(\alpha\), is the angle at which the particle is launched relative to the horizontal. This angle affects the shape and distance of the projectile's path.
Interestingly, the horizontal range remains the same if the angle is changed to \(90^\circ - \alpha\). This is shown by the range formula:
\[R = \frac{u^2 \sin(2 \alpha)}{g}\]
Since \(\sin(2(90^\circ - \alpha)) = \sin(2\alpha)\), both angles produce the same range, creating symmetrical projectile paths.
Horizontal Range
The horizontal range is the total horizontal distance the particle travels before landing. It's affected by both the initial speed and the angle of projection. The formula to calculate the range is:
\[R = \frac{u^2 \sin(2 \alpha)}{g}\]
In this equation, \(g\) represents the acceleration due to gravity. The horizontal range is maximized when the angle of projection is \(45^\circ\), demonstrating the importance of both velocity components in determining range.
Time of Flight
The time of flight is the total time the particle spends in the air. This can be calculated using the formula:
\[T = \frac{2u \sin \alpha}{g}\]
From the horizontal range and time relation, we can express time of flight in another form by:
\[T = \frac{R}{u \cos \alpha}\]
Understanding the time of flight helps in predicting where and when the particle will land, showing the journey of the projectile in both vertical and horizontal directions.
Maximum Height
The maximum height is the highest point reached by the particle during its motion. This value depends on the initial speed and the angle of projection:
\[H = \frac{u^2 \sin^2 \alpha}{2g}\]
At this peak, the vertical component of the velocity is zero, and only the horizontal component persists. Calculating the maximum height is crucial for understanding the vertex of the projectile's parabolic trajectory.
Kinematics Equations
Kinematics equations describe the motion of the projectile, considering the initial speed, angle, acceleration due to gravity, and time. The relevant kinematic equations for projectile motion include:
  • Horizontal Motion: \(x = u \cos \alpha \cdot t\)
  • Vertical Motion: \(y = u \sin \alpha \cdot t - \frac{1}{2} g t^2\)
These equations enable us to resolve many aspects of projectile motion and simplify the calculation for range, time of flight, and height.

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Most popular questions from this chapter

A particle is projected from the origin \(\mathrm{O}\) with velocity \(V\) at an elevation \(\theta\) to the horizontal. Show that its height \(y\) above \(\mathrm{O}\) when it has travelled a distance \(x\) horizontally is given by \(y=x \tan \theta-\frac{g x^{2} \sec ^{2} \theta}{2 V^{2}}\) A ball thrown from \(O\) with speed \(1400 \mathrm{~cm} / \mathrm{s}\) is caught at a point \(P\), which is \(1000 \mathrm{~cm}\) horizontally from \(O\) and \(187.5 \mathrm{~cm}\) above the level of 0 . Find the two possible angles of projection. If the ball is thrown from \(\mathrm{O}\) with the same initial speed to pass through a point \(562.5 \mathrm{~cm}\) vertically above \(P\), show that there is only one possible angle of projection. (U of L)

A vertical mast OP is \(h \mathrm{ft}\). high and stands with \(\mathrm{O}\) on the horizontal ground. (a) Two particles are projected simultaneously from P in the same vertical plane with the same speed, but with different angles of projection. Show that the distance between the particles increases untformly with time. (b) If a particle is projected vertically upwards from \(\mathrm{O}\) with velocity \(V\) and a second particle is projected at the same instant from P with velocity \(V\) and angle of projection \(\theta\) show that they are at their shortest distance apart after time \(h /(2 \mathrm{~V})\), and find this shortest distance. (Cambridge)

Two equal particles are projected at the same instant from points \(\mathrm{A}\) and \(\mathrm{B}\) at the same level, the first from A towards \(\mathrm{B}\) with velocity \(u\) at \(45^{\circ}\) above \(\mathrm{AB}\), and the second from B towards A with velocity \(v\) at \(60^{\circ}\) above BA. If the particles collide directly when each reaches its greatest height, find the ratio \(v^{2}: u^{2}\) and prove that \(u^{2}=g a(3-\sqrt{3})\), where \(a\) is the distance \(\mathrm{AB}\). After the collision the first particle falls vertically. Show that the coefficient of restitution between the particles is \((\sqrt{3}-1)(\sqrt{3}+1)\). (J.M.B)

Two particles are projected simultaneously from the same point with angles of projection \(\alpha\) and \(\beta\) and initial speeds \(u\) and \(v .\) Show that at any time during their flight the line joining them is inclined at \(\arctan \left(\frac{u \sin \alpha-v \sin \beta}{u \cos \alpha-v \cos \beta}\right)\) to the hori- zontal.

A particle is projected from a point \(O\) to hit a target which is level with \(O\). Find the two possible angles of projection. (a) The target is \(100 \mathrm{~m}\) from \(\mathrm{O}\). (b) The mass of the projectile is \(0.005 \mathrm{~kg}\). (c) The initial speed of the projectile is \(35 \mathrm{~ms}^{-1}\).
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