Oblique Projectile: Comprehensive NEET Physics Notes

1. Introduction to Oblique Projectile

An oblique projectile refers to a type of motion where an object is projected at an angle other than 90° to the horizontal. This motion can be analyzed as a combination of both horizontal and vertical movements, influenced by gravity, resulting in a parabolic trajectory. Understanding this concept is crucial for NEET aspirants, as it is a common topic in physics problems involving motion.

2. Basic Concepts

When a projectile is launched with an initial velocity $v_0$ at an angle ${\theta }_0$ with respect to the horizontal, its motion is governed by two separate components:

• Horizontal motion: Occurs with a constant velocity (no acceleration in the absence of air resistance).
• Vertical motion: Influenced by gravity, exhibiting constant acceleration downward.

2.1 Initial Velocity Components

The initial velocity $v_0$ can be broken down into horizontal and vertical components:

• Horizontal Component: $v_{0x}=v_0\cos {\theta }_0$
• Vertical Component: $v_{0y}=v_0\sin {\theta }_0$

2.2 Equations of Motion

For oblique projectile motion, the position, velocity, and acceleration of the projectile at any time $t$ are described as follows:

Horizontal Direction:

• Displacement: $x=v_{0x}t=(v_0\cos {\theta }_0)t$
• Velocity: $v_x=v_{0x}=v_0\cos {\theta }_0$

Vertical Direction:

• Displacement: $y=v_{0y}t-\frac{1}{2}g{t}^2=(v_0\sin {\theta }_0)t-\frac{1}{2}g{t}^2$
• Velocity: $v_y=v_{0y}-gt=v_0\sin {\theta }_0-gt$

Resultant Velocity at any time t: $v=\sqrt{v_x^2+v_y^2}$

Did You Know?

Galileo was the first to describe the independence of horizontal and vertical motions, which form the foundation of projectile motion analysis.

2.3 Path of the Projectile

The projectile follows a parabolic trajectory described by the equation: $y=x\tan {\theta }_0-\frac{g{x}^2}{2v_0^2{\cos }^2{\theta }_0}$ This equation is critical for understanding how the horizontal and vertical components interact to form the curved path.

Real-life Application:

The oblique projectile motion is observed in sports such as cricket or football, where the ball follows a parabolic path when kicked or thrown.

3. Key Parameters

3.1 Time of Flight (T)

The total time for which the projectile remains in the air is given by: $T=\frac{2v_0\sin {\theta }_0}{g}$

3.2 Maximum Height (H)

The highest vertical position reached by the projectile: $H=\frac{v_0^2{\sin }^2{\theta }_0}{2g}$

3.3 Horizontal Range (R)

The total horizontal distance covered by the projectile: $R=\frac{v_0^2\sin 2{\theta }_0}{g}$ The range is maximum when ${\theta }_0=45^{\circ }$.

Common Misconception:

Many students assume that the horizontal component of velocity changes during the flight. However, in the absence of air resistance, it remains constant.

4. Analysis and Problem-Solving Strategy

NEET Problem-Solving Strategy

1. Resolve the initial velocity into horizontal and vertical components using trigonometric functions.
2. Apply equations of motion separately for the horizontal and vertical components.
3. Utilize key formulas such as time of flight, maximum height, and range to solve problems efficiently.

NEET Tip:

Always start by identifying known values and variables, then use the appropriate formula. This method saves time and reduces errors during the exam.

5. Visual Aids

Suggested Diagrams

1. A diagram showing the projectile's trajectory with labeled components (initial velocity, horizontal range, maximum height, and angle of projection).
2. Graphs showing how the horizontal and vertical components change over time.

Incorporating these visual aids will significantly enhance understanding and make the concepts more memorable.

6. Practice Questions with Solutions

Question 1

An object is launched at an angle of 30° with a velocity of 25 m/s. Find the time of flight, maximum height, and horizontal range. Take $g=9.8{\text{m/s}}^2$.

Solution

1. Time of Flight: $T=\frac{2v_0\sin {\theta }_0}{g}=\frac{2\times 25\times \sin 30^{\circ }}{9.8}=\frac{25}{9.8}\approx 2.55\text{s}$
2. Maximum Height: $H=\frac{v_{}}{2g}$