Problem 20 - Master Resource Book in JEE Main Physics

Question

The trajectory of a projectile in vertical plane is \(y=a x-b x^{2}\), where \(a\) and \(b\) are constants and \(x\) and \(y\) are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are (a) \(\frac{b^{2}}{4 b}, \tan ^{-1}(b)\) (b) \(\frac{a^{2}}{b}, \tan ^{-1}(2 b)\) (c) \(\frac{a^{2}}{4 b}, \tan ^{-1}(a)\) (d) \(\frac{2 a^{2}}{b}, \tan ^{-1}(a)\)

Step-by-Step Solution

Answer

Option (c): \(\frac{a^2}{4b}, \tan^{-1}(a)\).

Step 1: Identify Maximum Height

The maximum height is attained when \(\frac{dy}{dx} = 0\). Given the equation \(y = ax - bx^2\), we first find the derivative: \(\frac{dy}{dx} = a - 2bx\). Setting this to zero for maximum height gives: \[ a - 2bx = 0 \]\[ a = 2bx \] \[ x = \frac{a}{2b} \].

Step 2: Calculate Maximum Height

Substitute the value of \(x = \frac{a}{2b}\) back into the equation \(y = ax - bx^2\):\[ y = a\left(\frac{a}{2b}\right) - b\left(\frac{a}{2b}\right)^2 \]\[ y = \frac{a^2}{2b} - \frac{ba^2}{4b^2} \]Combine terms:\[ y = \frac{a^2}{2b} - \frac{a^2}{4b} = \frac{2a^2}{4b} - \frac{a^2}{4b} = \frac{a^2}{4b} \].

Step 3: Find Angle of Projection

The angle of projection \(\theta\) is the angle at which the projectile is launched, determined by the initial slope or tangent of the curve y = ax - bx^2 at \(x = 0\). Here, the tangent at this point is the initial slope \(a\), so:\[ \theta = \tan^{-1}(a) \].

Step 4: Determine Correct Option

The maximum height calculated is \(\frac{a^2}{4b}\) and the angle of projection is \(\tan^{-1}(a)\). Therefore, the correct option is:(c) \(\frac{a^2}{4b}, \tan^{-1}(a)\).

Key Concepts

Maximum Height of Projectile

In projectile motion, the maximum height is the highest point reached by the projectile during its flight. To find this, we look where the vertical velocity component equals zero. For the equation given, which is a quadratic function, the derivative represents the rate of change of height with respect to horizontal distance. At the peak, this derivative is zero because the projectile stops moving upward and starts descending. To solve for this, we differentiate the trajectory equation, which is denoted by:- Given: \( y = ax - bx^2 \)- Derivative: \( \frac{dy}{dx} = a - 2bx \)We set the derivative \( \frac{dy}{dx} \) equal to zero to find the horizontal distance \( x \) where the highest point occurs:- \( a - 2bx = 0 \)- Solve: \( x = \frac{a}{2b} \)By substituting \( x \) back into the original equation, we calculate the maximum height \( y \). Replacing gives:- \( y = a\left(\frac{a}{2b}\right) - b\left(\frac{a}{2b}\right)^2 \) - Simplified: \( y = \frac{a^2}{4b} \)Thus, the maximum height is \( \frac{a^2}{4b} \). This emphasizes the influence of both constants \( a \) and \( b \) in determining this peak height.

Angle of Projection

The angle of projection, or launch angle, is the angle at which the projectile is fired with respect to the horizontal plane. This angle is crucial because it determines the trajectory's path and ultimately affects the range and height of the projectile. In our given scenario, this angle can be determined by examining the slope of the trajectory at the origin point (i.e., when \( x = 0 \)).To find this initial slope, we refer again to the derivative of the trajectory equation \( y = ax - bx^2 \). The slope \( a \) at \( x = 0 \), is essentially the tangent of the angle of projection, formulated as:- Initial slope: \( \tan(\theta) = a \)Hence, the angle of projection \( \theta \) is:- \( \theta = \tan^{-1}(a) \)Understanding this angle helps in designing the initial conditions of launch, ensuring the projectile covers the desired trajectory efficiently. It connects directly to the factors influencing the launch, making \( \theta \, (angle of projection) \) a vital part of analyzing projectile motion.

Equation of Trajectory

The equation of a projectile's trajectory describes its path and is essential for predicting the movement of objects in a gravitational field when air resistance is negligible. In our task, the trajectory equation is provided as \( y = ax - bx^2 \), where \( a \) and \( b \) are constants.Breaking this down, this equation is a standard form of a parabola, opening downwards, which is typical for projectile motion under Earth's gravity. Important features of this parabolic trajectory include:

• Vertex: Represents the highest point, which we found to be \( \frac{a^2}{4b} \).
• Axis of Symmetry: Vertical line through the vertex, \( x = \frac{a}{2b} \), indicating where the projectile achieves its maximum height.
• Direction: Since it simplifies to a negative squared term (-bx²), this implies a downward opening.

This equation effectively models the motion by considering both horizontal motion (influenced by \( x \)) and vertical drop (where \( x^2 \) comes into play because of gravity). Understanding this relationship aids in applying concepts to real-world physics problems, like predicting how far a ball will travel or how high it will climb.