Problem 33 - Master Resource Book in JEE Main Physics

Question

For a projectile thrown into space with a speed \(v\), the horizontal range is \(\frac{\sqrt{3} v^{2}}{2 g} .\) The vertical range is \(\frac{v^{2}}{8 g}\). The angle which the projectile makes with the horizontal initially is (a) \(15^{*}\) (b) \(30^{*}\) (c) \(45^{\circ}\) [d) \(60^{\circ}\)

Step-by-Step Solution

Answer

The angle is \(30^{\circ}\).

Step 1: Recognize the Known Ranges

We are given two formulas for the ranges: the horizontal range \( R_h = \frac{\sqrt{3} v^{2}}{2 g} \) and the vertical range \( R_v = \frac{v^{2}}{8 g} \). We must determine the initial angle from these.

Step 2: Understand Projectile Motion Equations

For a projectile motion, the horizontal range \( R_h \) can be given by \( R_h = \frac{v^2 \sin(2\theta)}{g} \), and the vertical range (maximum height) \( H \) is given by \( H = \frac{v^2 \sin^2(\theta)}{2g} \).

Step 3: Match Formulas with Given Data

Using the equation for horizontal range, we have: \( \frac{v^2 \sin(2\theta)}{g} = \frac{\sqrt{3} v^{2}}{2 g} \). Similarly, for vertical range, \( \frac{v^2 \sin^2(\theta)}{2g} = \frac{v^{2}}{8 g} \).

Step 4: Solve for \( \theta \) from Horizontal Range Equation

Cancelling \(v^2/g\) from both sides in the horizontal range equation yields \( \sin(2\theta) = \frac{\sqrt{3}}{2} \). The angle \(2\theta\) which satisfies this is \(60^{\circ}\), so \(\theta = 30^{\circ}\).

Step 5: Verify \( \theta \) with Vertical Range Equation

Plug in \( \theta = 30^{\circ} \) into the vertical range formula to check: \( \sin^2(30^{\circ}) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) and \( \frac{v^2 \frac{1}{4}}{2g} = \frac{v^2}{8g} \). It matches, confirming \(\theta = 30^{\circ}\) is correct.

Key Concepts

Horizontal Range

In projectile motion, the horizontal range refers to the total horizontal distance a projectile covers from the launch point until it lands again on the same level. It is influenced by the initial speed and launch angle of the projectile.
For a projectile launched at an angle \( \theta \) with initial speed \( v \), the horizontal range \( R_h \) can be calculated using the equation:

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

Where:

• \( v \) is the initial velocity,
• \( g \) is the acceleration due to gravity, approximately \( 9.81 \, m/s^2 \),
• \( \theta \) is the launch angle.

The double angle \( \sin(2\theta) \) indicates that the range depends on the sine of two times the angle. This fact helps us find the optimal angles that maximize the range, specifically around \( 45^{\circ} \). However, the given problem uses a specific situation to calculate a known range.

Vertical Range

The vertical range or maximum height of a projectile describes how high the projectile travels before descending back. This happens when the vertical component of its velocity becomes zero at the peak of its trajectory.
For a projectile thrown with an initial speed \( v \) at an angle \( \theta \), the formula for maximum height \( H \) is:

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

Where:

• \( \sin^2(\theta) \) is the square of the sine of the angle, indicating that height is maximized when \( \sin(\theta) \) is large,
• \( g \) is the acceleration due to gravity.

In the specific problem, the vertical range matches the given formula \( \frac{v^2}{8g} \), confirming the derived angle. Understanding the vertical range helps in determining how high the projectile will go, contributing to solving projectile motion problems.

Projectile Angle

The projectile angle \( \theta \) is crucial because it determines the projectile's range and maximum height. The angle sets the initial direction of the projectile's velocity vector.
To find the optimal trajectory for a desired distance or height, you need to choose an appropriate angle, often requiring the application of trigonometric identities.
The typical method of determining projectile angle in problems involves solving equations using known conditions like specific horizontal or vertical ranges.
This problem uses given range equations to solve for \( \theta \). In the calculated solution, matching the known formulas \( \sin(2\theta) = \frac{\sqrt{3}}{2} \) gives a solution angle of \( \theta = 30^{\circ} \). Choosing the projectile angle wisely ensures the maximum efficiency of projectile motion.

Trigonometric Functions

Trigonometric functions play an integral role in projectile motion. They relate the angle, initial speed, and the projectile's trajectory.
The primary functions involved usually include sine \( \sin \) and cosine \( \cos \). These functions help decompose initial velocity into horizontal and vertical components:

• The horizontal component \( v \cos(\theta) \)
• The vertical component \( v \sin(\theta) \)

The trigonometric identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \) simplifies calculations of range and angle, allowing one to find efficient solutions for projectile motion paths.
Understanding and applying these functions accurately can determine key characteristics of the projectile's path, enhancing problem-solving skills in physics.