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A projectile has a time of flight \(T\) and range \(R\). If the time of flight is doubled, what happens to the range? a. \(R / 4\) b. \(R / 2\) c. \(2 R\) d. \(4 R\)

Short Answer

Expert verified
Doubling the time of flight results in the range becoming four times as large, \(4R\). Therefore, the correct answer is d: \(4R\).

Step by step solution

01

Understand the relationship between time of flight and range

The range \( R \) of a projectile is given by the formula \( R = v_i \cdot T \cdot \cos(\theta) \), where \( v_i \) is the initial velocity, \( T \) is the time of flight, and \( \theta \) is the angle of projection. The time of flight \( T \) is determined by \( T = \frac{2v_i \sin(\theta)}{g} \), where \( g \) is the acceleration due to gravity.
02

Examine the effect of doubling the time of flight

Doubling the time of flight means the new time of flight \( T' = 2T \). According to the equation for \( T \), this would imply doubling the initial velocity \( v_i \), as \( T = \frac{2v_i \sin(\theta)}{g} \). No change in gravity (\( g \)) or angle (\( \theta \)) is assumed.
03

Calculate the effect on range

With \( v_i \) doubled, the new range \( R' = v_i' \cdot T' \cdot \cos(\theta) \), where \( v_i' = 2v_i \). Substituting \( T' = 2T \) and \( v_i' = 2v_i \) into the range formula, we find \( R' = 2v_i \cdot 2T \cdot \cos(\theta) = 4R \).
04

Select the correct answer

Based on the calculations, when the time of flight is doubled, the range quadruples. Therefore, the correct answer is option d: \( 4R \).

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Key Concepts

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

Time of Flight
Time of flight is the duration that a projectile spends in the air from the point of launch to the point it lands. It's a critical component in projectile motion.
Understanding time of flight helps you predict how long the projectile will be airborne.The formula for time of flight, given a projectile launched with an initial velocity \( v_i \) at an angle \( \theta \), is:
  • \( T = \frac{2v_i \sin(\theta)}{g} \)
Here, \( g \) represents the acceleration due to gravity, which is approximately 9.8 m/s² on Earth.
This formula indicates that time of flight depends on both the initial velocity and the angle of projection.Doubling the time of flight implies changes in the initial conditions of the launch, which affects other aspects like the range of the projectile.
Range of Projectile
The range of a projectile is the horizontal distance it covers during its flight. It represents the extent of the projectile's path measured along the x-axis.
This is especially important in applications such as sports, engineering, and warfare.The formula to calculate the range is:
  • \( R = v_i \cdot T \cdot \cos(\theta) \)
Here, \( v_i \) is the initial velocity, \( T \) is the time of flight, and \( \theta \) is the angle of projection.
The range depends on the initial velocity, angle, and time of flight.When the time of flight is doubled while maintaining the same other conditions, the range also changes significantly. As seen in the exercise solution, doubling the time of flight results in quadrupling the range, showing how interconnected these variables are.
Initial Velocity
Initial velocity is the speed at which a projectile is launched. It's crucial in determining how far and how long a projectile will travel.
In real-world scenarios, calculating initial velocity helps in making precise adjustments for desired outcomes.When a projectile is launched at an initial velocity \( v_i \), this velocity comprises two components:
  • Horizontal component: \( v_{i_x} = v_i \cdot \cos(\theta) \)
  • Vertical component: \( v_{i_y} = v_i \cdot \sin(\theta) \)
These components influence the projectile's range and time of flight.The exercise highlights that by doubling the time of flight, one might infer that the initial velocity is effectively doubled, due to the direct relationship established in the formula \( T = \frac{2v_i \sin(\theta)}{g} \). This change leads to an increase in range as well.
Angle of Projection
The angle of projection is the angle at which a projectile is launched relative to the horizontal axis. It significantly influences the path and outcome of the projectile's motion.
Understanding this angle aids in optimizing trajectories for maximum distance or specific targets.The effect of the angle is embedded within both time of flight and range formulas:
  • \( T = \frac{2v_i \sin(\theta)}{g} \)
  • \( R = v_i \cdot T \cdot \cos(\theta) \)
This angle determines how "high" or "flat" the projectile's path will be.A key takeaway is that while doubling time of flight can greatly increase the range, changes in the angle can further fine-tune these outcomes.
No change in angle was assumed in the exercise, meaning the existing trajectory and its impact on range remained consistent.

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Most popular questions from this chapter

The range \(R\) of projectile is same when its maximum heights are \(h_{1}\) and \(h_{2}\). What is the relation between \(R, h_{1}\), and \(h_{2} ?\) a. \(R=\sqrt{h_{1} h_{2}}\) b. \(R=\sqrt{2 h_{1} h_{2}}\) c. \(R=2 \sqrt{h_{1} h_{2}}\) d. \(R=4 \sqrt{h_{1} h_{2}}\)

Two balls \(A\) and \(B\) are thrown with speeds \(u\) and \(u / 2\), respectively. Both the balls cover the same horizontal distance before returning to the plane of projection. If the. angle of projection of ball \(B\) is \(15^{\circ}\) with the horizontal, then the angle of projection of Ais a. \(\sin ^{-1}\left(\frac{1}{8}\right)\) b. \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{8}\right)\). c. \(\frac{1}{3} \sin ^{-1}\left(\frac{1}{8}\right)\) d. \(\frac{1}{4} \sin ^{-1}\left(\frac{1}{8}\right)\)

The point from where a ball is projected is taken as the ori-gin of the coordinate axes. The \(x\) and \(y\) components of its displacement are given by \(x=6 t\) and \(y=8 t-5 t^{2} .\) What is the velocity of projection? a. \(6 \mathrm{~m} / \mathrm{s}\) b. \(8 \mathrm{~m} / \mathrm{s}\) c. \(10 \mathrm{~m} / \mathrm{s}\) d. \(14 \mathrm{~m} / \mathrm{s}\)

The angular velocity of a particle moving in a circle of radius \(50 \mathrm{~cm}\) is increased in 5 min from 100 revolutions per minute to 400 revolutions per minute. Find tangential acceleration of the particle. a. \(60 \mathrm{~m} / \mathrm{s}^{2}\) b. \(\pi / 30 \mathrm{~m} / \mathrm{s}^{2}\) c. \(\pi / 15 \mathrm{~m} / \mathrm{s}^{2}\) d. \(\pi / 60 \mathrm{~m} / \mathrm{s}^{2}\)

A shot is fired from a point at a distance of \(200 \mathrm{~m}\) from the foot of a tower \(100 \mathrm{~m}\) high so that it just passes over it horizontally. The direction of shot with horizontal is a. \(30^{\circ}\) b. \(45^{\circ}\) c. \(60^{\circ}\) d. \(70^{\circ}\)

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