Problem 36 - Master Resource Book in JEE Main Physics

Question

Two projectiles \(A\) and \(B\) are thrown with velocities \(v\) and \(\frac{v}{2}\) respectively. They have the same range. If \(B\) is thrown at an angle of \(15^{\circ}\) to the horizontal, A must have been thrown at an angle (a) \(\sin ^{-1}\left(\frac{1}{16}\right)\) (b) \(\sin ^{-1}\left(\frac{1}{4}\right)\) (c) \(2 \sin ^{-1}\left(\frac{1}{4}\right)\) (d) \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{8}\right)\)

Step-by-Step Solution

Answer

The angle for projectile A is \( \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \), option (d).

Step 1: Understand the Projectile Range Formula

The range for a projectile fired with velocity \( v \) at an angle \( \theta \) is given by the formula \( R = \frac{v^2 \sin(2\theta)}{g} \), where \( g \) is the acceleration due to gravity.

Step 2: Establish Equation for Projectile B

Projectile B has a velocity of \( \frac{v}{2} \) and is thrown at an angle of \( 15^\circ \). Its range \( R_B \) is given by \[ R_B = \frac{\left(\frac{v}{2}\right)^2 \sin(30^\circ)}{g} = \frac{v^2 \sin(30^\circ)}{4g}. \]

Step 3: Establish Equation for Projectile A

Projectile A has a velocity \( v \) and an unknown angle \( \theta \). Its range \( R_A \) is given by \( R_A = \frac{v^2 \sin(2\theta)}{g} \).

Step 4: Set Ranges Equal to Each Other

Since it is given that both projectiles have the same range \( R_A = R_B \), set the equations equal: \[ \frac{v^2 \sin(2\theta)}{g} = \frac{v^2 \sin(30^\circ)}{4g}. \]

Step 5: Simplify the Equation

Cancel \( v^2/g \) from both sides of the equation to get: \[ \sin(2\theta) = \frac{1}{4}\sin(30^\circ). \]

Step 6: Solve for \( \theta \)

We know \( \sin(30^\circ) = \frac{1}{2} \), hence the equation becomes \( \sin(2\theta) = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \). Thus, \( 2\theta = \sin^{-1}\left(\frac{1}{8}\right) \), which means \( \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \).

Step 7: Choose the Correct Option

By comparing \( \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \) to the given options, we select \( (d) \frac{1}{2} \sin^{-1}\left(\frac{1}{8}\right) \) as the correct answer.

Key Concepts

Projectile Range Formula

Understanding the projectile range formula is fundamental when dealing with projectile motion problems. The range of a projectile refers to the horizontal distance it travels while in motion. This is an essential concept in physics as it relates to how far an object will land when projected at a certain velocity and angle. The formula for the range of a projectile launched with initial velocity \( v \) at an angle \( \theta \) is given by:

• \( R = \frac{v^2 \sin(2\theta)}{g} \)

Here, \( g \) is the acceleration due to gravity, typically approximated as \( 9.81 \text{ m/s}^2 \) on the surface of the Earth. In this formula, notice how the range is influenced by the square of the velocity and a trigonometric function of the angle.
This means both the velocity and the angle at which the object is projected are crucial for determining how far it will travel. The \( \sin(2\theta) \) term reveals that the most effective angle for maximizing range is not simply \( 45^{\circ} \), but dependent on the sine of twice the angle of projection. Understanding this formula allows students to predict the behavior of a projectile across various velocities and angles.

Trigonometric Functions in Physics

Trigonometric functions are vital tools in physics for relating angles and distances. They are particularly useful in analyzing projectile motion. In the context of projectiles, functions like sine, cosine, and tangent frequently appear, as they relate the components of vectors such as velocity and force.
For the current problem, sine functions are used to determine the range of projectiles fired at certain angles. If a projectile is launched at an angle \( \theta \) to the horizontal:

• The vertical component of velocity is affected by \( \sin(\theta) \).
• The horizontal component is influenced by \( \cos(\theta) \).

The use of \( \sin(2\theta) \) in the range formula shows how trigonometric identities can simplify and solve problems involving angles. Additionally, understanding the inverse sine function, \( \sin^{-1} \), is critical for solving equations where the angle itself is unknown. In this problem, solving the equation using inverse sine was necessary to determine the correct angle at which projectile A was launched. Recognizing and manipulating these functions is a key problem-solving skill in physics.

Physics Problem Solving Techniques

Solving physics problems often involves systematic techniques and understanding of fundamental principles. Here are some strategies that can help when tackling problems involving projectile motion:

• **Identifying Known and Unknown Variables:** First, list what values are given in the problem and what you need to find. Identify the equations that relate these variables.
• **Using Appropriate Formulas:** Understanding the right formulas, like the projectile range formula, and how to apply them correctly is crucial.
• **Simplification:** Break down complex equations into simpler steps. In this problem, cancelling common terms and reducing the equation simplified the process.
• **Trigonometric Considerations:** Use trigonometric identities and functions to handle angles and components of vectors efficiently.
• **Verification:** Once you have a solution, verify it by checking against the original problem. Does it satisfy all conditions posed by the problem statement?

By following these techniques, students can methodically approach and solve physics problems. Such strategies build strong problem-solving skills that extend beyond academic settings, fostering logical thinking and precision.