/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 \- A body is thrown from the sur... [FREE SOLUTION] | 91Ó°ÊÓ

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\- A body is thrown from the surface of the earth with velocity \(\left(g R_{e}\right) / 2\), where \(R_{e}\) is the radius of the earth at some angle from the vertical. If the maximum height reached by the body is \(R / 4\), then the angle of projection with the vertical is (1) \(\sin ^{-1}\left(\frac{\sqrt{5}}{4}\right)\) (2) \(\cos ^{-1}\left(\frac{\sqrt{5}}{4}\right)\) (3) \(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) (4) none of these

Short Answer

Expert verified
The correct angle is (1) \(\sin^{-1}\left(\frac{\sqrt{5}}{4}\right)\).

Step by step solution

01

Understand the Problem

The problem states that a body is projected at an angle with a given initial velocity and we need to find the angle at which it was projected given the maximum height reached.
02

Know the given values

You are given the initial velocity of the projectile as \(v = \frac{g R_e}{2}\) and the maximum height \(H = \frac{R}{4}\).

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Key Concepts

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

Initial Velocity
The initial velocity of a projectile is the speed at which it is launched into motion. In this problem, the initial velocity is given as \(v = \frac{gR_e}{2}\).
  • \(g\) represents the gravitational acceleration, which is approximately 9.81 m/s² on Earth's surface.
  • \(R_e\) is the Earth's radius, which is about 6,371 kilometers.
This formula combines gravitational force and the scale of the Earth to define a specific launch speed. Initial velocity is crucial because it determines how high and far an object will travel.In projectile motion problems, understanding initial velocity helps calculate other attributes like range and time of flight. In this exercise, it's essential to correctly interpret the initial value to solve for the angle of projection.
Maximum Height
The maximum height is the highest vertical position reached by a projectile during its motion. In this scenario, it is given as \(H = \frac{R}{4}\), where \(R\) could represent an assumed or calculated value dependent on context, possibly linked to Earth in symbolic expressions.The maximum height is reached when the vertical component of the velocity becomes zero due to gravitational pull acting in the opposite direction.
  • At max height, vertical velocity \(v_y = 0\).
  • It is influenced by the vertical component of the initial velocity and gravity.
To solve for this maximum height, the formula \(H = \frac{v_y^2}{2g}\) can be used, showing how vertical velocity and gravitational pull impact height. Accurately calculating maximum height allows determination of crucial angles and projectile trajectory characteristics.
Angle of Projection
The angle of projection is the angle at which a projectile is launched relative to a reference plane, such as the ground or the vertical. This angle is vital in determining the flight path and maximum height of the projectile. For this exercise, the challenge is to find at what angle the projectile was launched with respect to the vertical to achieve a certain height. Understanding the angle of projection involves:
  • Analyzing its effect on both horizontal range and vertical height.
  • Using trigonometric relationships to determine how vertical and horizontal velocity components are influenced by this angle.
The connection between angle, initial velocity, and height is evident as both horizontal and vertical components of velocity affect how high and far the projectile will travel.
Vertical Motion
Vertical motion refers to the component of projectile motion that deals with movement along the vertical axis, primarily influenced by gravity.In projectile motion, the vertical motion determines how high and how quickly the object reaches maximum height. It includes:
  • The initial vertical component of velocity \(v_{y0} = v \sin(\theta)\).
  • The effect of gravity, a constant 9.81 m/s² downward.
  • The change in velocity and position, determined by kinematic equations.
For example, the time taken to reach the maximum height can be calculated using \(t = \frac{v_{y0}}{g}\). By understanding vertical motion, one can solve for the maximum height and total time in the air, both critical in accurately predicting projectile behavior.

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Most popular questions from this chapter

The distance between the centres of the Moon and the earth is \(D .\) The mass of the earth is 81 times the mass of the Moon. At what distance from the centre of the earth, the gravitational force will be zero? (1) \(\frac{D}{2}\) (2) \(\frac{2 D}{3}\) (3) \(\frac{4 D}{3}\) (4) \(\frac{9 D}{10}\)

At perihelion, the mechanical energy of Pluto's orbit has (1) its maximum value (2) its minimum value (3) the same value as at every other point in the orbit (4) the value which depends on the sense of rotation

A double star consists of two stars having masses \(M\) and \(2 M\). The distance between their centres is equal to \(r .\) They revolve under their mutual gravitational interaction. Then, which of the following statements are not correct? (1) Heavier star revolves in orbit of radius \(2 r / 3\) (2) Both the stars revolve with the same speed, period of which is equal to \(\left(2 \pi / r^{3}\right)\left(2 G M^{2} / 3\right)\) (3) Kinetic energy of the heavier star is twice that of the other star. (4) None of the above

Which of the following are not correct? (1) The escape velocity for the Moon is \(6 \mathrm{~km} \mathrm{~s}^{-1}\). (2) The escape velocity from the surface of Moon is \(v\). The orbital velocity for a satellite to orbit very close to the surface of Moon is \(v / 2\). (3) When an earth satellite is moved from one stable orbit to a further stable orbit, the gravitational potential energy increases. (4) The orbital velocity of a satellite revolving in a circular path close to the planet is independent of the density of the planet.

A space vehicle approaching a planet has a speed \(v\), when it is very far from the planet. At that moment tangent of its trajectory would miss the centre of the planet by distance \(R\). If the planet has mass \(M\) and radius \(r\), what is the smallest value of \(R\) in order that the resulting orbit of the space vehicle will just miss the surface of the planet? (1) \(\frac{r}{v}\left[\nu^{2}+\frac{2 G M}{P}\right]^{\frac{1}{2}}\) (2) \(v r\left[1+\frac{2 G M}{P}\right]\) (3) \(\frac{r}{v}\left[v^{2}+\frac{2 G M}{r}\right]\) (4) \(\frac{2 G M v}{r}\)

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