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Chapter 5: Problem 25

Two particles are projected from the same point with the same speed at different angles \(\theta_{1}\) and \(\theta_{2}\) to the horizontal They have the same range. Their times of flight are \(t_{1}\) and \(t_{2}\), respectively. Then choose the correct option(s). (1) \(\theta_{1}=90-\theta_{2}\) (2) \(\frac{t_{1}}{t_{2}}=\tan \theta_{2}\) (3) \(\frac{t_{1}}{\sin \theta_{1}}=\frac{t_{2}}{\sin \theta_{2}}\) (4) \(\frac{t_{1}}{t_{2}}=\tan \theta_{1}\)

Short Answer

Expert verified
Options 1 and 3 are correct.

Step by step solution

01

Recall the formula for range

The range of a projective motion is given by the formula \( R = \frac{v^2 \sin 2\theta}{g} \), where \( v \) is the speed of projection, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity. Since both particles have the same range, \( \sin 2\theta_1 = \sin 2\theta_2 \). This implies that either \( \theta_1 = \theta_2 \) or \( \theta_1 + \theta_2 = 90^\circ \).
02

Analyze option 1

Option 1 states that \( \theta_1 = 90^\circ - \theta_2 \). From the conclusion in Step 1, this is a valid condition because if \( \sin 2\theta_1 = \sin 2\theta_2 \), one possibility is \( \theta_1 + \theta_2 = 90^\circ \). Therefore, option 1 is correct.
03

Recall the formula for time of flight

The time of flight for each particle can be calculated using \( t = \frac{2v \sin \theta}{g} \). For the two particles, their times of flight are \( t_1 = \frac{2v \sin \theta_1}{g} \) and \( t_2 = \frac{2v \sin \theta_2}{g} \).
04

Evaluate option 2

Option 2 states that \( \frac{t_1}{t_2} = \tan \theta_2 \). Using the expressions from Step 3, we have \( \frac{t_1}{t_2} = \frac{\sin \theta_1}{\sin \theta_2} \). This does not lead to \( \tan \theta_2 \) unless \( \theta_1 \) or \( \theta_2 \) is a specific value like 45 degrees, which is not generally true, so option 2 is incorrect.
05

Evaluate option 3

Option 3 states that \( \frac{t_1}{\sin \theta_1} = \frac{t_2}{\sin \theta_2} \). Substituting the expressions from Step 3, both sides become equal to \( \frac{2v}{g} \), hence this statement is true. Option 3 is correct.
06

Evaluate option 4

Option 4 states that \( \frac{t_1}{t_2} = \tan \theta_1 \). From Step 4, we found \( \frac{t_1}{t_2} = \frac{\sin \theta_1}{\sin \theta_2} \), which does not equate to \( \tan \theta_1 \) without specific angle values, thus making this option incorrect.

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Key Concepts

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

Range of Projectile
In projectile motion, the range is the total horizontal distance covered by the projectile. It is influenced by the initial speed, the angle of projection, and the acceleration due to gravity. The formula to calculate the range (\( R \)) is given by:\[ R = \frac{v^2 \sin 2\theta}{g} \]Where:
  • \( v \) is the initial speed of the projectile,
  • \( \theta \) is the angle of projection measured from the horizontal axis, and
  • \( g \) is the acceleration due to gravity, approximately 9.8 m/s虏 on Earth.
To achieve the same range with two different angles of projection, the angles must satisfy a critical relationship. Specifically, if two projectiles have the same initial speed and are projected with angles\( \theta_1 \)and\( \theta_2 \), then:\[ \theta_1 + \theta_2 = 90^\circ \]This finding reveals that the angles are complementary, meaning they add up to 90 degrees. This is critical when solving projectile motion problems. It indicates that one angle projects the same distance as another when each angle is on either side of the maximum range angle.
Time of Flight
The time of flight describes how long a projectile is in motion from the point it leaves the ground until it returns to the same horizontal level. The time of flight depends on the initial speed and the angle of projection. It can be calculated for a projectile using the formula:\[ t = \frac{2v \sin \theta}{g} \]Where the variables are defined similarly as in the range formula:
  • \( v \) is the initial velocity,
  • \( \theta \) is the angle of projection, and
  • \( g \) is the gravitational acceleration.
The total time of flight increases with larger angles of projection because the projectile travels a greater vertical distance. While solving exercises, if two projectiles launched with the same speed have the exact range, their time of flight relates by their angle projections. Fractionally,\( \frac{t_1}{\sin \theta_1} \)will equal\( \frac{t_2}{\sin \theta_2} \), signifying a deeper connection between the time of flight and the angles employed. This coherence aids in solving problems with different angle and time configurations.
Angle of Projection
The angle at which a projectile is launched highly affects its trajectory and range. The angle of projection (\( \theta \)) determines how a projectile moves through the air and achieves its range.To understand this, it鈥檚 crucial to note:
  • As the angle of projection increases, the projectile will cover a higher arc.
  • For a fixed speed, there is an optimal angle that maximizes the range. This maximum range is typically obtained at an angle of\( 45^\circ \).
  • Angles producing the same range must be complementary, meaning,\( \theta_1 + \theta_2 = 90^\circ \).
Analyzing projectile problems often involves determining the right angles for the desired outcomes. By focusing on understanding the relationships between the launch angles,\( \theta_1 \)and\( \theta_2 \), students can better predict how changes in angles affect the motion and effectiveness of the projectile. This knowledge is crucial not only in physics problems but in various practical applications where trajectory analysis is essential.

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Most popular questions from this chapter

A particle is thrown at time \(t=0\) with a velocity of \(10 \mathrm{~ms}^{-1}\) at an angle \(60^{\circ}\) with the horizontal from a point on an inclined plane, making an angle of \(30^{\circ}\) with the horizontal. The time when the velocity of the \(79 .\) projectile becomes parallel to the incline is (1) \(\frac{2}{\sqrt{3}} \mathrm{~s}\) (2) \(\frac{1}{\sqrt{3}} \mathrm{~s}\) (3) \(\sqrt{3} \mathrm{~s}\) (4) \(\frac{1}{2 \sqrt{3}} \mathrm{~s}\)

Two inclined planes \(O A\) and \(O B\) having inclination (with borizontal) \(30^{\circ}\) and \(60^{\circ}\), respectively, intersect each other at \(O\) as shown in figure. A particle is projected from point \(P\) with velocity \(u=10 \sqrt{3} \mathrm{~ms}^{-1}\) along a direction perpendicular to plane \(O A\). If the particle strikes plane \(O B\) perpendicularly at \(Q\), calculate The velocity with which particle strikes the plane \(O B\), (1) \(15 \mathrm{~ms}^{-1}\) (2) \(30 \mathrm{~ms}^{\prime}\) (3) \(20 \mathrm{~ms}^{-1}\) (4) \(10 \mathrm{~ms}^{-1}\)

A bead is free to slide down on a smooth wire rightly stretched between points \(A\) and \(B\) on a vertical circle of radius \(10 \mathrm{~m}\). Find the time taken by the bead to reach point \(B\), if the bead slides from rest from the highest point \(A\) on the circle.

Two ports, \(A\) and \(B\), on a North-South line are separated by a \(\pi_{\text {le }}\) of width \(D\). The river flows east with speed \(u\). A boat crosses the river starting from port \(A .\) The speed of the boat relative to the river is \(v\). Assume \(v=2 u\). Suppose the boat wants to cross the river from \(A\) to the other side in the shortest possible time. Then what should be the direction of the velocity of boat relative to river? (1) \(30^{\circ}\) west of north (2) \(30^{\circ}\) east of north (3) \(60^{\circ}\) west of north (4) along north

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 \(\tan \alpha ?\)
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