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Chapter 5: Problem 5

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 ]

Short Answer

Expert verified
The initial velocity \( v_0 \) is approximately 11.4 ms\(^{-1}\).

Step by step solution

01

Breakdown the Known Values

We are given that the maximum height, \( H \), is 5 m and the radius of curvature, \( r \), at the highest point is 3 m. We need to find the initial velocity, \( v_0 \). Also, use \( g = 10 \ \text{ms}^{-2} \).
02

Relation between Radius of Curvature and Velocity

The radius of curvature, \( r \), at the highest point of the trajectory is given by:\[r = \frac{v_0^2 \cos^2\theta}{g}.\]Substituting the known values, we have:\[3 = \frac{v_0^2 \cos^2\theta}{10}.\]This simplifies to:\[v_0^2 \cos^2\theta = 30.\]
03

Relation between Maximum Height and Velocity

The maximum height attained by the projectile is given by the formula:\[H = \frac{v_0^2 \sin^2\theta}{2g}.\]Plug in the known values:\[5 = \frac{v_0^2 \sin^2\theta}{20}.\]This simplifies to:\[v_0^2 \sin^2\theta = 100.\]
04

Simplifying using Trigonometric Identity

From steps 2 and 3, we have:\[v_0^2 \cos^2\theta = 30\]\[v_0^2 \sin^2\theta = 100\]We know that \(\sin^2\theta + \cos^2\theta = 1\). Adding these equations gives:\[v_0^2 (\cos^2\theta + \sin^2\theta) = 130\]Thus, we find:\[v_0^2 = 130\]
05

Solve for Initial Velocity

Taking the square root of both sides, we have:\[v_0 = \sqrt{130}.\]Given \(\sqrt{80}\) is approximately 9, estimate:\[\sqrt{130} \approx 11.4 \].Therefore, \(v_0 \approx 11.4 \ \text{ms}^{-1}\).

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Key Concepts

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

Understanding Initial Velocity in Projectile Motion
When a body is propelled into the air, the concept of *initial velocity*, often denoted by \(v_0\), plays a crucial role in determining the trajectory of the object. In projectile motion, the initial velocity is the speed at which the object is launched. It can be dissected into two components due to the angle of launch:
  • Horizontal Component: \(v_0 \cos \theta\)
  • Vertical Component: \(v_0 \sin \theta\)
These components help in calculating various aspects of the projectile鈥檚 path.In the given exercise, we are tasked with finding the initial velocity knowing the maximum height and radius of curvature. This aspect shows how initial velocity interplays with other parameters in projectile motion. The initial velocity is a foundational factor since it influences how high and how far the projectile will travel.
Exploring Maximum Height in Projectile Motion
The *maximum height* of a projectile refers to the highest vertical position it reaches during its flight. At this point, all the initial vertical velocity has been converted into potential energy, and the object begins its descent.Calculating maximum height can be done using:\[H = \frac{v_0^2 \sin^2\theta}{2g}\].In this relation:
  • \(v_0\) represents the initial velocity.
  • \(\sin \theta\) relates to the vertical component of the initial speed.
  • \(g\) is the acceleration due to gravity, often approximated as \(9.8 \mathrm{ms}^{-2}\) but given as \(10 \mathrm{ms}^{-2}\) in this problem.
Given that our exercise provides a maximum height, you can substitute this into the formula to solve for parts of the initial velocity, further helping in triangulating solutions in motion problems.
Understanding Radius of Curvature at Maximum Height
The *radius of curvature* in projectile motion provides insights into the path鈥檚 shape at a specific point. At the peak of the trajectory, this concept captures how sharply the projectile turns around. Mathematically expressed, it is shown as: \[r = \frac{v_0^2 \cos^2\theta}{g}\]Here, the radius of curvature depends heavily on:
  • The horizontal velocity component \(v_0 \cos \theta\), which is the velocity perpendicular to the resultant path at the peak.
  • The gravitational acceleration \(g\), adding downward force.
In the problem, knowing the radius of curvature allows us to form an equation involving the initial velocity. This is incredibly useful as it gives another relationship that must be satisfied alongside our height equation, resulting in a complete understanding of the projectile鈥檚 motion, linking back to the initial velocity computations. Understanding this concept helps forecast how smoothly or sharply an object will move through its trajectory.

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Most popular questions from this chapter

A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of \(2 \mathrm{~m} / \mathrm{s} .\) Now the wind screen is placed at angle \(\alpha\) with the vertical such that the rain drops falling vertically downwards with velocity \(6 \mathrm{~m} / \mathrm{s}\) strike the wind screen perpendicularly. What is the value of \(\tan \alpha ?\)

Two balloons are simultaneously released from two buildings \(A\) and \(B\). Balloon from A rises with constant velocity \(10 \mathrm{~ms}^{-1}\), While the other one rises with constant velocity of \(20 \mathrm{~ms}^{-1}\). Due to wind the balloons gather horizontal velocity \(V_{x}=0.5 y\), where \(y\) is the height from the point of release. The buildings are at a distance of \(250 \mathrm{~m}\) and after some time \(t\) the balloons collide. Choose the correct option(s). (1) \(t=5 \mathrm{sec}\) (2) difference in height of buildings is \(100 \mathrm{~m}\) (3) difference in height of buildings is \(50 \mathrm{~m}\) (4) \(t=10 \mathrm{sec}\)

A particle is projected with a certain velocity at an angle \(\alpha\) above the horizontal from the foot of an inclined plane of inclination \(30^{\circ}\). If the particle strikes the plane normally. then \(\alpha\) is equal to (1) \(30^{\circ}+\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) (2) \(45^{\circ}\) (3) \(60^{\circ}\) (4) \(30^{\circ}+\tan ^{-1}(2 \sqrt{3})\)

The speed of a projectile at its highest point is \(v_{1}\) and at the point half the maximum height is \(v_{2}\). If \(\frac{v_{1}}{v_{2}}=\sqrt{\frac{2}{5}}\), then fin? the angle of projection. (1) \(45^{\circ}\) (2) \(30^{\circ}\) (3) \(37^{\circ}\) (4) \(60^{\circ}\)

A student throws soft balls out of the window at different angles to the horizontal. All soft balls have the same inity speed \(v=10 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\). It turns out that all soft balls' landing velocities make angles \(30^{\circ}\) or greater with the horizontal Find the height \(h\) (in \(\mathrm{m}\) ) of the window above the ground.
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