Problem 54 - Master Resource Book in JEE Main Physics

Question

A particle is projected from the ground with an initial speed of \(v\) at an angle \(\theta\) with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is (a) \(\frac{v}{2} \sqrt{1+2 \cos ^{2} \theta}\) (b) \(\frac{v}{2} \sqrt{1+\cos ^{2} \theta}\) (c) \(\frac{v}{2} \sqrt{1+3 \cos ^{2} \theta}\) (d) \(v \cos \theta\)

Step-by-Step Solution

Answer

The average velocity is \( \frac{v}{2} \sqrt{1+2 \cos ^{2} \theta} \), which corresponds to option (a).

Step 1: Find the Highest Point of the Trajectory

The highest point of the trajectory can be found by focusing only on the vertical component of the velocity. The vertical component of the initial velocity is \( v \sin \theta \). At the highest point, the vertical velocity component becomes zero due to gravity's opposition, leading to the equation: \( v_y = v\sin \theta - gt = 0 \), where \( t \) is the time to reach the highest point and \( g \) is the acceleration due to gravity. Solving for \( t \), we find \( t = \frac{v \sin \theta}{g} \).

Step 2: Determine Horizontal Distance Covered

The horizontal component of the velocity remains constant at \( v \cos \theta \). The horizontal distance \( x \) covered while reaching the highest point can be calculated as \( x = v \cos \theta \times t \). Substituting the value of \( t \) from Step 1, we get \( x = v \cos \theta \times \frac{v \sin \theta}{g} = \frac{v^2 \sin \theta \cos \theta}{g} \).

Step 3: Calculate Vertical Distance to Highest Point

The vertical distance \( y \) to the highest point can be calculated using the formula for vertical displacement: \( y = v \sin \theta \times t - \frac{1}{2}gt^2 \). Substituting the time \( t = \frac{v \sin \theta}{g} \), we get \( y = \frac{v^2 \sin^2 \theta}{2g} \).

Step 4: Find Magnitude of Displacement to Highest Point

The displacement from the start to the highest point is the vector sum of the horizontal and vertical distances found. This gives: \( \text{displacement} = \sqrt{x^2 + y^2} \). Substituting \( x = \frac{v^2 \sin \theta \cos \theta}{g} \) and \( y = \frac{v^2 \sin^2 \theta}{2g} \), we find: \( \text{displacement} = \frac{v^2}{g} \sqrt{\frac{\sin^2 \theta \cos^2 \theta}{g} + \frac{\sin^4 \theta}{4g^2}} \). Simplifying this gives us \( \text{displacement} = \frac{v^2 \sin^2 \theta}{g}\sqrt{\cos^2 \theta + \frac{\sin^2 \theta}{4}} \).

Step 5: Calculate Average Velocity

The average velocity is given by the total displacement (as found in Step 4) over total time, which is \( \frac{\text{displacement}}{t} \). Substituting the value of \( t = \frac{v \sin \theta}{g} \), we get \( \text{average velocity} = \frac{v^2 \sin^2 \theta}{g} \times \frac{g}{v \sin \theta} = \frac{v}{2} \sqrt{1 + 2 \cos^2 \theta} \).

Step 6: Match with Given Options

Compare the derived expression for the average velocity with the provided options. Our expression \( \frac{v}{2} \sqrt{1 + 2 \cos^2 \theta} \) matches with option (a), confirming our solution.

Key Concepts

Average Velocity

In physics, average velocity is determined by dividing the total displacement by the total time taken. It’s important to differentiate between speed and velocity. While speed is a scalar quantity measuring how fast an object is moving, velocity is a vector quantity, which means it also includes direction.
\[ \text{Average velocity} = \frac{\text{Total displacement}}{\text{Total time}} \]
The average velocity in projectile motion gives us an understanding of the nett movement over a time interval, rather than moment-by-moment speed. For our problem, we calculated average velocity between the projection point and the highest point of the trajectory, which led us to the expression \( \frac{v}{2} \sqrt{1 + 2 \cos^2 \theta} \). This tells us that the average velocity depends on the cosine of the launch angle \( \theta \), illustrating the impact of horizontal motion in the calculation.
Understanding this concept helps in predicting motion trajectories in real-life scenarios based on initial velocity and angle parameters.

Vertical Motion

Vertical motion in projectile motion focuses on how gravity affects a projectile's vertical component. When a particle is projected with an initial velocity, the vertical component is defined as \( v \sin \theta \). Gravity acts downward, opposing this initial velocity, and eventually reduces it to zero at the peak of the trajectory.
At the highest point, the time taken for vertical deceleration is given by:
\[ t = \frac{v \sin \theta}{g} \]
Here, \( g \) represents the acceleration due to gravity. This concept is critical because it helps determine the time a projectile spends moving upward until it begins its descent.
Once the projectile reaches its highest point, it begins to fall back down, and gravity continues to affect it by increasing the downward speed.

Horizontal Displacement

Horizontal displacement in projectile motion looks at how far the projectile travels along the horizontal axis while moving upward to the highest point. Unlike vertical motion, horizontal motion isn't affected by gravity, meaning that the horizontal component of velocity, \( v \cos \theta \), remains constant throughout the flight.
The horizontal distance covered or displacement to the highest point is calculated using:
\[ x = v \cos \theta \times t \]
Substituting the time \( t \) we found for vertical motion, we arrive at:
\[ x = \frac{v^2 \sin \theta \cos \theta}{g} \]
This expression shows the influence of both initial speed and the launch angle on the projectile's displacement. Understanding horizontal displacement is crucial in predicting where the projectile will land and how far it will travel.

Trajectory Analysis

Trajectory analysis in projectile motion integrates both vertical and horizontal motion to map the projectile's path through space. The trajectory is essentially the curved path a projectile follows under the influence of both its initial projection and gravity.
The total displacement from the projection point to the highest point is a combination of both horizontal and vertical displacements. It can be computed as:
\[ \text{Displacement} = \sqrt{x^2 + y^2} \]
Where \( x \) is the horizontal displacement and \( y \) is the vertical displacement. By analyzing the trajectory, one can determine the impact of different angles and speeds on the path of the projectile.
In real life, this analysis helps engineers and scientists predict how an object will behave when launched, making it possible to design various solutions and understand phenomena – from sports mechanics to space vehicle launches.