Class 11 > Unit # 02: Kinematics > Derivations of Projectile Motion

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Unit # 2: Kinematics — Mathematical Derivations of Projectile Motion

Lecture Overview

This lecture from Talha's Physics Academy covers the essential step-by-step mathematical derivations for projectile motion parameters in Class 11 Physics (Unit 2: Kinematics), including time to reach maximum height, total time of flight, maximum height attained, horizontal range, and condition for maximum range.

Source Video: Derivation of Projectile Motion | Total time of flight, Maximum Height , Horizontal Range

Introduction

A projectile is fired with initial velocity \(v_0\) making an angle \(\theta\) with the horizontal surface. Below are the step-by-step derivations for its key parameters:

A) Time to Reach Maximum Height

Figure: Ascent to Maximum Height

Visual breakdown showing the projectile moving upward until its vertical velocity component becomes zero at maximum height.

Given Data:

• Initial Velocity = \(v_{oy} = v_0 \sin\theta\)
• Final Velocity = \(v_y = 0\)
• Time = \(T\)
• Acceleration = \(-g\)

Using the first equation of motion:

$$v_f = v_i + at$$

$$v_y = v_{oy} + (-g)T$$

$$0 = v_0 \sin\theta - gT$$

$$gT = v_0 \sin\theta$$

$$T = \frac{v_0 \sin\theta}{g}$$

B) Total Time of Flight

Figure: Total Flight Path

Complete trajectory showing starting and ending points at the same horizontal reference level.

Total time of flight is denoted by \(T'\):

$$T' = 2T$$

$$T' = \frac{2v_0 \sin\theta}{g}$$

C) Maximum Height Attained

Given Data:

• Distance = Height = \(y = h_{\text{max}}\)
• Initial Velocity = \(v_{oy} = v_0 \sin\theta\)
• Acceleration = \(a = -g\)
• Time = \(T = \frac{v_0 \sin\theta}{g}\)

Using the second equation of motion:

$$S = v_i t + \frac{1}{2}at^2$$

$$h_{\text{max}} = v_{oy}T + \frac{1}{2}(-g)T^2$$

$$h_{\text{max}} = (v_0 \sin\theta)\left(\frac{v_0 \sin\theta}{g}\right) - \frac{1}{2}g\left(\frac{v_0 \sin\theta}{g}\right)^2$$

$$h_{\text{max}} = \frac{v_0^2 \sin^2\theta}{g} - \frac{v_0^2 \sin^2\theta}{2g}$$

$$h_{\text{max}} = \frac{2v_0^2 \sin^2\theta - v_0^2 \sin^2\theta}{2g}$$

$$h_{\text{max}} = \frac{v_0^2 \sin^2\theta}{2g}$$

D) Horizontal Range

The maximum horizontal distance traveled by a projectile is called range.

Given Data:

• Distance = \(X = R\)
• Velocity = \(v_{ox} = v_0 \cos\theta\)
• Time = \(T'\)

Using uniform velocity formula for horizontal motion:

$$S = v \times t$$

$$R = v_{ox} \times T'$$

$$R = (v_0 \cos\theta)\left(\frac{2v_0 \sin\theta}{g}\right)$$

$$R = \frac{v_0^2 (2 \cos\theta \sin\theta)}{g}$$

Using trigonometric identity \(\sin(2\theta) = 2 \sin\theta \cos\theta\):

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$

E) Maximum Range

Horizontal range is given as:

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$

Above expression shows that, for a constant velocity of projection (\(v_0\)) and gravitational acceleration (\(g\)), the horizontal range depends on the factor \(\sin(2\theta)\) and it will be maximum at the maximum value of \(\sin(2\theta)\). The maximum value of sine is \(1\).

$$\sin(2\theta) = 1$$

$$2\theta = \sin^{-1}(1)$$

$$2\theta = 90^\circ$$

$$\theta = 45^\circ$$

It shows that, "when a projectile is projected at \(45^\circ\), its horizontal range will be maximum."

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