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Chapter 5: Problem 2

A projectile can have same range \(R\) for two angles of projection. It \(t_{1}\) and \(t_{2}\) are the times of flight in the two cases, then what is the product of two times of flight? (1) \(t_{1} t_{2} \propto R^{2}\) (2) \(t_{1} t_{2} \propto R\) (3) \(t_{1} r_{2} \propto \frac{1}{R}\) (4) \(t_{1} t_{2} \propto \frac{1}{R^{2}}\)

Short Answer

Expert verified
The correct answer is (2): \(t_{1} t_{2} \propto R\).

Step by step solution

01

Understanding the Problem

The problem states that a projectile can achieve the same range when projected at two different angles. We need to determine the relationship between the product of the times of flight for these angles and the range.
02

Recall Projectile Motion Equations

For a given projection velocity, the range of a projectile fired at angle \( \theta \) can be given by \( R = \frac{v^2 \sin 2\theta}{g} \). The time of flight is \( T = \frac{2v \sin \theta}{g} \).
03

Different Angles for Same Range

For the same range, the angles of projection \( \theta \) and \( 90^\circ - \theta \) will both achieve the range \( R \). Therefore, we consider the times of flight for angles \( \theta \) and \( 90^\circ - \theta \).
04

Calculate Times of Flight

Find \( t_1 \) and \( t_2 \) using the formula for time of flight: \( t_1 = \frac{2v \sin \theta}{g} \) and \( t_2 = \frac{2v \sin (90^\circ - \theta)}{g} = \frac{2v \cos \theta}{g} \).
05

Compute the Product of Times of Flight

The product \( t_1 t_2 = \left( \frac{2v \sin \theta}{g} \right) \left( \frac{2v \cos \theta}{g} \right) = \frac{4v^2 \sin \theta \cos \theta}{g^2} \).
06

Use Trigonometric Identity and Range Equation

Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we modify the equation: \( t_1 t_2 = \frac{2v^2 \sin 2\theta}{g^2} = \frac{2gR}{g^2} = \frac{2R}{g} \).
07

Final Calculation

From the above step, \( t_1 t_2 = \frac{2R}{g} \). This confirms that \( t_1 t_2 \propto R \).

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Key Concepts

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

Time of Flight
The time of flight of a projectile is a crucial aspect in projectile motion. It refers to the total time a projectile remains in the air from the moment it is launched until it lands. This time depends on several factors, such as the initial velocity and the angle of projection.

In general, the formula for the time of flight for a projectile launched with an initial velocity \(v\) at an angle \(\theta\) to the horizontal is given by:
  • \( T = \frac{2v \sin \theta}{g} \)
where \(g\) is the acceleration due to gravity, approximately \(9.8 \text{m/s}^2\). The sine component \( \sin \theta \) indicates that the time of flight increases with steeper angles until \(\theta = 90^\circ\), where it reaches a maximum.

For the same range but different angles, the time of flight varies. For instance, angles \(\theta\) and \(90^\circ - \theta\) give different times, denoted as \(t_1\) and \(t_2\) respectively. Despite varying flight times, when we multiply these times, the dependency on the range \(R\) becomes evident, as demonstrated by:
  • \( t_1 t_2 = \frac{2R}{g} \)
This means the product of the times is directly proportional to the range, showing that even with different flight durations, the relationship with range remains tight.
Range of Projectile
Understanding the range of a projectile helps in determining how far the object will travel horizontally. The range is influenced primarily by the initial speed and the angle of projection. Depending on these factors, the projectile can cover vast distances or fall short.

The range for a given launch speed \(v\) and angle \(\theta\) is expressed by:
  • \( R = \frac{v^2 \sin 2\theta}{g} \)
Interestingly, this formula simplifies when the projectile is fired at angles \(\theta\) and \(90^\circ - \theta\), which produce identical ranges. Here, the range simplifies our understanding of symmetry in projectile motion. The range is maximized at a \(45^\circ\) angle, which spreads the energy evenly between horizontal and vertical motions.

When assessing the range equation and its symmetry, one can recognize that the highest range is achieved when the projectile is neither too horizontal nor too vertical. This balanced approach ensures the greatest horizontal displacement.
Angles of Projection
The angle of projection is the angle at which an object is launched relative to the horizontal. This angle plays a pivotal role in defining the trajectory, time of flight, and ultimately the range of the projectile.

With the equation \( R = \frac{v^2 \sin 2\theta}{g} \), you can see why angles \(\theta\) and \(90^\circ - \theta\) produce the same range. The term \(\sin 2\theta\) ensures that this property holds due to the characteristics of the sine function, which remains the same for complimentary angles adding to \(90^\circ\).

The choice of angle also influences the path or trajectory shape. Lower angles result in flatter trajectories with less time airborne, while higher angles yield taller, more parabolic paths with longer flight durations. Some key points about angles and their effects:
  • **45掳 Angle:** Produces the highest range for a given speed.
  • **0掳 and 90掳 Angles:** Result in minimal or zero ranges.
  • **Symmetrical Ranges:** Achieved at angles \(\theta\) and \(90^\circ - \theta\).
Understanding which angle to choose and its effects helps in strategic planning where projectile motion is involved.

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Most popular questions from this chapter

A projectile has a time of flight \(T\) and range \(R\). If the time of flight is doubled, keeping the angle of projection same what happens to the range? (1) \(R / 4\) (2) \(\mathrm{R} / 2\) (3) \(2 \mathrm{R}\) (4) \(4 \mathrm{R}\)

A ball is projected from the ground at angle \(\theta\) with the horizontal. After \(1 \mathrm{~s}\), it is moving at angle \(45^{\circ}\) with the horizontal and after \(2 \mathrm{~s}\) it is moving horizontally. What is the velocity of projection of the ball? (1) \(10 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(20 \sqrt{3} \mathrm{~m} \mathrm{~s}^{-1}\) (3) \(10 \sqrt{5} \mathrm{~ms}^{-1}\) (4) \(20 \sqrt{2} \mathrm{~ms}^{-1}\)

Two bullets are fired horizontally with different velocities from the same height. Which will reach the ground first? (1) Slower one (2) Faster one (3) Both will reach simultaneously (4) Cannot be predicted

A particle moves along positive branch of the curre \(y=\frac{x}{2}\), where \(x=\frac{t^{3}}{3}, x\) and \(y\) are measured in meters an \(t\) in seconds, then (1) The velocity of particle at \(t=1 \mathrm{~s}\) is \(\hat{i}+\frac{1}{2} \hat{j}\). (2) The velocity of particle at \(t=1 \mathrm{~s}\) is \(\frac{1}{2} \hat{i}+\hat{j}\). (3) The acceleration of particle at \(t=2 \mathrm{~s}\) is \(2 \hat{i}+\hat{j}\). (4) The acceleration of particle at \(t=2 \mathrm{~s}\) is \(\hat{i}+2 \hat{j}\).

Rain appears to fall vertically to a man walking at \(3 \mathrm{~km} \mathrm{~h}^{-1}\), but when he changes his speed to double, the rain appears to fall at \(45^{\circ}\) with vertical. Study the following statements and find which of them are correct. i. Velocity of rain is \(2 \sqrt{3} \mathrm{~km} \mathrm{~h}^{-1}\) ii. The angle of fall of rain (with vertical) is $$ \theta=\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iii. The angle of fall of rain (with vertical) is $$ \theta=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right) \text {. } $$ iv. Velocity of rain is \(3 \sqrt{2} \mathrm{kmh}^{-1}\). (I) 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.
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