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A body is projected upwards with a velocity \(u\). It passes through a certain point above the ground after \(t_{1}\). The time after which the body passes through the same point during the return journey is a. \(\left(\frac{u}{g}-t_{\mathrm{I}}^{2}\right)\) b. \(2\left(\frac{u}{g}-t_{1}\right)\) c. \(3\left(\frac{u^{2}}{g}-t_{1}\right)\) d. \(3\left(\frac{u^{2}}{g^{2}}-t_{1}\right)\)

Short Answer

Expert verified
The correct answer is b: \(2\left(\frac{u}{g}-t_1\right)\).

Step by step solution

01

Understanding the problem context

We are dealing with a problem in classical mechanics where a body is projected upwards. Important variables are: initial velocity \(u\), gravitational acceleration \(g\), and a specific time \(t_1\) when the body passes a certain point during its upward journey. We need to find the time it takes to pass back through the same point on its downward journey.
02

Determine the time of ascent

The formula for the time to reach the maximum height is given by \( t_{ ext{ascent}} = \frac{u}{g} \). This formula arises because the velocity of the body becomes zero at the peak of its trajectory.
03

Time to reach the particular point during ascent

We know \(t_1\) is the time taken to reach a certain point during ascent. It usually means \(t_1\leq \frac{u}{g}\) since the body has not yet reached its maximum height.
04

Calculate the time for the round trip to the same point

The time to reach the same point on its descent can be derived using symmetry of motion. The total time taken to reach this point again after ascending for \(t_1\) is \(2\left(\frac{u}{g}\right) - t_1\) due to the symmetry of parabolic trajectory under constant acceleration.
05

Selecting the correct answer

From the calculation made in the previous step, the time of descent to the same point is expressed as \(2\left(\frac{u}{g} - t_{1}\right)\). Hence, the correct option is **b.**

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Key Concepts

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

Classical Mechanics
In classical mechanics, the study of motion and forces is fundamental for understanding how objects behave. This branch of physics provides a framework for analyzing the movement of bodies under the influence of various forces. In the case of projectile motion, classical mechanics gives us the tools to dissect the path and behavior of objects, considering factors like initial velocity, angle of launch, and gravitational pull.
In this exercise, we've tackled a projectile problem where a body is projected upwards. Applying classical mechanics allows us to predict the body's trajectory, the point it reaches, and the time it takes to travel along its path. These solutions rely on well-defined laws and formulas, such as when an object will stop ascending and start descending, helping us understand the complete motion cycle of the projectile.
Time of Ascent
Time of ascent is a key concept in understanding projectile motion, especially when a body is launched upwards. It refers to the time taken by the body to reach its peak point, where its velocity becomes zero before it starts descending.
The time of ascent can be calculated using the formula:\[ t_{ascent} = \frac{u}{g} \] Where:
  • \( u \): Initial velocity of the object
  • \( g \): Gravitational acceleration
Once the object reaches its maximum height, the time before it descends again is crucial for predicting its course, as it illustrates the motion symmetry in projectile paths. Understanding this helps in solving for the total time to hit a certain repeat point on its descent.
Gravitational Acceleration
Gravitational acceleration is a constant value that significantly impacts the motion of any object under its influence. On Earth, it is approximately \(9.81 \text{ m/s}^2\), pulling objects toward the center of the planet.
When dealing with projectile motion, gravitational acceleration determines how quickly an object climbs and descends. It plays a pivotal role in both the ascent and descent calculations. During ascent, gravity decelerates the object until it halts at its peak, and subsequently, it accelerates the object as it falls back down.
For our exercise, understanding that gravity equally affects the body during its trip explains why the times of ascent and descent to a corresponding height are related. This grasp of gravitational acceleration is crucial for mastering how forces influence moving bodies.
Symmetry of Parabolic Motion
The symmetry of parabolic motion is a fascinating feature of projectile paths. It stems from the constant gravitational force acting throughout the object's journey, resulting in a symmetric path characterized by equal ascent and descent times about the peak.
When a body is projected upwards, the motion can be viewed as a parabola. The path mirrors itself at the highest point, meaning the time to reach any specific altitude during the ascent is mirrored precisely on the descent.
The exercise utilizes this property to calculate the total time for the projectile to return to a specific point after reaching it initially during ascent. This symmetry simplifies computing projectile paths and demonstrating how uniform gravitational forces produce predictable and balanced motions.

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Most popular questions from this chapter

A parachutist drops first freely from an aeroplane for \(10 \mathrm{~s}\) and then parachute opens out. Now he descends with a net retardation of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). If he bails out of the plane at a height of \(2495 \mathrm{~m}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), his velocity on reaching the ground will be a. \(5 \mathrm{~m} / \mathrm{s}\) b. \(10 \mathrm{~m} / \mathrm{s}\) c. \(15 \mathrm{~m} / \mathrm{s}\) d. \(20 \mathrm{~m} / \mathrm{s}\)

A train is moving at a constant speed \(V\). Its driver obscrves another train in front of him on the same track and moving in the same dircction with constant speed \(v\). If the distance between the trains be \(x\), what should be the minimum retardation of the train so as to avoid collision? a. \(\frac{(V+v)^{2}}{x}\) b. \(\frac{(V-v)^{2}}{x}\) c. \(\frac{(V+v)^{2}}{2 x}\) d. \(\frac{(V-v)^{2}}{2 x}\)

A particle is dropped from rest from a large height. Assume \(g\) to be constant throughout the motion. The time taken by it to fall through successive distances of \(1 \mathrm{~m}\) each will be a. all equal, being equal to \(\sqrt{2 / g}\) second b. in the ratio of the square roots of the integers \(1,2,3, \ldots\) c. in the ratio of the difference in the square roots of the integers, i.e., \(\sqrt{1},(\sqrt{2}-\sqrt{1}),(\sqrt{3}-\sqrt{2}),(\sqrt{4}-\sqrt{3}), \ldots\) d. in the ratio of the reciprocals of the square roots of the integers, i.e., \(\frac{1}{\sqrt{1}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{3}}, \ldots\)

Check up only the correct statement in the following. a. A body has a constant velocity and still it can have a varying speed. b. A body has a constant speed but it can have a varying velocity. c. A body having constant speed cannot have any acceleration. d. None of these.

The average velocity of a body moving with uniform acceleration after travelling a distance of \(3.06 \mathrm{~m}\) is \(0.34 \mathrm{~ms}^{-1} .\) If the change in velocity of the body is \(0.18 \mathrm{~ms}^{-1}\) during this time, its uniform acceleration is a. \(0.01 \mathrm{~m} / \mathrm{s}^{2}\) b. \(0.02 \mathrm{~m} / \mathrm{s}^{2}\) c. \(0.03 \mathrm{~m} / \mathrm{s}^{2}\) d. \(0.04 \mathrm{~m} / \mathrm{s}^{2}\)

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