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Chapter 5: Problem 5

A particle is projected with a velocity \(v\) so that its range on a horizontal plane is twice the greatest height attained. If \(g\) is acceleration due to gravity, then its range is (1) \(\frac{4 v^{2}}{5 g}\) (2) \(\frac{4 g}{5 v^{2}}\) (3) \(\frac{4 v^{3}}{5 g^{2}}\) (4) \(\frac{4 v}{5 g^{2}}\)

Short Answer

Expert verified
The range is \(\frac{4v^2}{5g}\), which is option (1).

Step by step solution

01

Understanding the Range and Maximum Height

For a projectile motion, the range, \(R\), and the maximum height, \(H\), are given by the formulas \[ R = \frac{v^2 \sin 2\theta}{g} \]and\[ H = \frac{v^2 \sin^2 \theta}{2g} \]where \(\theta\) is the angle of projection. We need to express \(R = 2H\).
02

Substitute and Equate

Substitute the expressions for \(R\) and \(H\) in the equation \(R = 2H\):\[ \frac{v^2 \sin 2\theta}{g} = 2 \cdot \frac{v^2 \sin^2 \theta}{2g} \]This simplifies to:\[ \sin 2\theta = \sin^2 \theta \]
03

Simplify using Trigonometric Identity

Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]Divide both sides by \(\sin \theta\):\[ 2 \cos \theta = \sin \theta \]
04

Solve for \(\theta\)

Express \(\cos \theta\) in terms of \(\sin \theta\):\[ \cos \theta = \frac{\sin \theta}{2} \]Square both sides and use \(\sin^2 \theta + \cos^2 \theta = 1\):\[ \cos^2 \theta = \frac{\sin^2 \theta}{4} \]\[ \sin^2 \theta + \frac{\sin^2 \theta}{4} = 1 \]Solving gives:\[ \frac{5 \sin^2 \theta}{4} = 1 \]\[ \sin^2 \theta = \frac{4}{5} \]\[ \sin \theta = \frac{2}{\sqrt{5}} \]
05

Calculate the Range

Now, substitute \(\sin 2\theta = 2 \sin \theta \cos \theta = \sin^2 \theta\) and \(\sin 2\theta = \frac{4}{5}\) into the range formula:\[ R = \frac{v^2 \cdot \frac{4}{5}}{g} = \frac{4v^2}{5g} \]
06

Match with the Given Options

Compare the calculated range \(\frac{4v^2}{5g}\) with the options provided. The answer matches the first option: \(\frac{4v^2}{5g}\).

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Key Concepts

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

Range of Projectile
When studying projectile motion, one of the key aspects is understanding the range of a projectile. The range (0R0) is defined as the horizontal distance covered by a projectile. It is influenced by the initial velocity (0v0) and the angle of projection (0\theta0). To calculate the range, you can use the formula: \[ R = \frac{v^2 \sin 2\theta}{g} \]Here, 0g0 stands for the acceleration due to gravity, which typically is 09.81 \text{ m/s}^20.
  • The range is maximum when the projectile is launched at a 045^0 angle since 0\sin 90^0 (which is 02\theta0 when 0\theta = 45^0) is 010.
  • The direction and magnitude of the projectile's initial velocity significantly impact the trajectory.
  • Range is directly proportional to the square of the velocity, meaning any change in initial speed notably affects the projectile's horizontal distance.
In the problem, the range equates to twice the maximum height, helping to further analyze and solve the scenario.
Maximum Height of Projectile
Understanding the maximum height (0H0) that a projectile can attain is crucial in analyzing projectile motion. The maximum height refers to the peak point in the projectile's trajectory, where vertical velocity becomes zero momentarily before descending. To find the maximum height, use this formula: \[ H = \frac{v^2 \sin^2 \theta}{2g} \]Some critical points about the maximum height are:
  • The angle 0\theta0 of the projectile heavily influences the maximum height. As the angle increases from 0 to 90 degrees, 0\sin \theta0 initially increases, but the angle for greatest height is less than that of range.
  • Maximizing the initial vertical component (0v \sin \theta0) increases 0H0.
  • At maximum height, all the projectile's kinetic energy is converted into potential energy.
In the exercise scenario, setting this height in relation to the range was key to finding the solution by mathematical manipulation.
Trigonometric Identities in Physics
Trigonometric identities play a crucial role in solving projectile motion problems, as they help simplify complex expressions. Some common trigonometric identities used include:
  • \( \sin 2\theta = 2 \sin \theta \cos \theta \)
  • \( \sin^2 \theta + \cos^2 \theta = 1 \)
These identities are vital in converting between different trigonometric functions and solving equations related to projectile motion.For instance, in the exercise presented, the identity 0\sin 2\theta = 2 \sin \theta \cos \theta0 was used to express equations in terms of a single trigonometric function. By applying algebraic manipulation, such as dividing or squaring both sides, solving for one specific variable becomes manageable. Understanding how these identities operate allows students to derive results efficiently and accurately, especially when linking physical scenarios with mathematical concepts.Mastering these trigonometric concepts is beneficial not only for physics problems but also greatly aids proficiency in mathematics overall.

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Most popular questions from this chapter

Raindrops are hitting the back of a man walking at a speed of \(5 \mathrm{kmh}^{-1}\). If he now starts running in the same direction with a constant acceleration, the magnitude of the velocity of the rain with respect to him will (1) gradually increase (2) gradually decrcase (3) first decrease then increase (4) first increase then decrease

Two ports, \(A\) and \(B\), on a North-South line are separated by a \(\pi_{\text {le }}\) of width \(D\). The river flows east with speed \(u\). A boat crosses the river starting from port \(A .\) The speed of the boat relative to the river is \(v\). Assume \(v=2 u\). The boat crosses the river from \(A\) to the other side in shortest possible time, then how far is the boat from the port \(B\) after crossing the river (1) \(D / \sqrt{2}\) (2) \(\sqrt{2} D\) (3) \(2 D\) (4) \(D / 2\)

A man holds an umbrella at \(30^{\circ}\) with the vertical to keep himself dry. He, then, runs at a speed of \(10 \mathrm{~ms}^{-1}\), and find the rain drops to be hitting vertically. Study the following statements and find the correct options. i. Velocity of rain w.r.t. Earth is \(20 \mathrm{~m} \mathrm{~s}^{-1}\) ii. Velocity of rain w.r.t. man is \(10 \sqrt{3} \mathrm{~ms}^{-1}\) iii. Velocity of rain w.r.t. Earth is \(30 \mathrm{~ms}^{-1}\) iv. Velocity of rain w.r.t. man is \(10 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (1) Statements (i) and (ii) are correct. (2) Statements (i) and (iii) are correct. (3) Statements (iii) and (iv) are correct. (4) Statements (ii) and (iv) are correct.

Points \(A\) and \(C\) are on the horizontal ground and \(A\) and \(B\) are in same vertical plane at a distance of \(1500 \mathrm{~m}\). Simultaneously bullets are fired from \(A, B\) and \(C\) and they collide at \(O\). The bullet at \(B\) is fired at an angle of \(30^{\circ}\) with horizontal towards the ground at velocity \(100 \mathrm{~m} / \mathrm{s}\). The bullet at \(C\) is projected vertically upward at velocity of \(100 \mathrm{~m} / \mathrm{s}\). The bullet projected from \(A\) reaches its maximum height at \(O\). Find the time in which bullets will collide (seconds): (1) 10 (2) 15 (3) 20 (4) 25

An aeroplane is flying vertically upwards. When it is at a height of \(1000 \mathrm{~m}\) above the ground and moving at a speed of \(367 \mathrm{~m} / \mathrm{s}\), a shot is fired at it with a speed of \(567 \mathrm{~m} \mathrm{~s}^{-1}\) from a point directly below it. What should be the acceleration of aeroplane so that it may escape from being hit? (1) \(>5 \mathrm{~ms}^{-2}\) (2) \(>10 \mathrm{~ms}^{-2}\) (3) \(<10 \mathrm{~ms}^{-2}\) (4) Not possible
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