Projectile Motion Formulas: Derivations

Hello! Welcome to your fourth lesson in our study of two-dimensional kinematics.

In our previous session, we successfully derived the trajectory equation for a projectile, $y (x)$ , and rigorously proved its parabolic nature by transforming it into the standard form of a parabola. This gave us a static picture of the projectile's path.

Today, we will analyze the key dynamic features of this path. Our learning outcome is to derive, with rigorous proofs, the formulas for maximum height, time of flight, and horizontal range of a projectile. We will start from the fundamental kinematic equations and derive these important results from first principles, focusing on the physical conditions that define each quantity. This is a crucial step in mastering projectile motion for JEE-level problems.

Let's begin by recalling the component-wise equations of motion for a projectile launched from the origin ( $x_{0} = 0, y_{0} = 0$ ) with initial velocity $v_{0}$ at an angle $θ$ to the horizontal:

Position:
- $x (t) = (v_{0} cos θ) t$
- $y (t) = (v_{0} sin θ) t - \frac{1}{2} g t^{2}$
Velocity:
- $v_{x} (t) = v_{0} cos θ$
- $v_{y} (t) = v_{0} sin θ - g t$

These equations will be our starting point.

1. Maximum Height ( $H$ )

The maximum height, $H$ , is the highest vertical position the projectile reaches. At the apex of the trajectory, the projectile's vertical motion momentarily ceases before it begins to descend.

Physical Condition and Derivation

The physical condition defining the apex is that the vertical component of velocity, $v_{y}$ , is zero.

Step 1: Find the time to reach maximum height ( $t_{p e ak}$ )
We use the vertical velocity equation and set $v_{y} (t) = 0$ :
$v_{y} (t_{p e ak}) = v_{0} sin θ - g t_{p e ak} = 0$
Solving for $t_{p e ak}$ , we get:
$t_{p e ak} = \frac{v _{0} s i n θ}{g}$

Step 2: Calculate the maximum height ( $H$ )
The maximum height is the vertical position $y$ at time $t_{p e ak}$ . We substitute this time into the vertical position equation:
$H = y (t_{p e ak}) = (v_{0} sin θ) t_{p e ak} - \frac{1}{2} g t_{p e ak}^{2}$
Now, substitute the expression for $t_{p e ak}$ :
$H = (v_{0} sin θ) (\frac{v _{0} s i n θ}{g}) - \frac{1}{2} g (\frac{v _{0} s i n θ}{g})^{2}$
Simplifying the terms:
$H = \frac{( v _{0} s i n θ ) ^{2}}{g} - \frac{1}{2} g (\frac{( v _{0} s i n θ ) ^{2}}{g ^{2}})$
$H = \frac{( v _{0} s i n θ ) ^{2}}{g} - \frac{( v _{0} s i n θ ) ^{2}}{2 g}$
Combining the terms gives us the final formula for maximum height:
$H = \frac{( v _{0} s i n θ ) ^{2}}{2 g}$

Alternative Derivation

We can also derive this result using the time-independent kinematic equation $v_{y}^{2} = v_{0 y}^{2} + 2 a_{y} s_{y}$ , which you derived in Module 2. At the maximum height, we have:

Final vertical velocity $v_{y} = 0$
Initial vertical velocity $v_{0 y} = v_{0} sin θ$
Vertical displacement $s_{y} = H$
Vertical acceleration $a_{y} = - g$

Substituting these into the equation:
$0^{2} = (v_{0} sin θ)^{2} + 2 (- g) H$
Solving for $H$ directly gives:
$H = \frac{( v _{0} s i n θ ) ^{2}}{2 g}$
This confirms our previous result and shows the consistency of the kinematic framework.

2. Time of Flight ( $T$ ) and Horizontal Range ( $R$ )

The time of flight, $T$ , is the total duration the projectile is in the air. The horizontal range, $R$ , is the total horizontal distance it travels during this time. For this derivation, we assume the projectile lands at the same vertical level from which it was launched.

Physical Condition and Derivation

The physical condition defining the end of the flight is that the projectile has returned to its initial launch height. Since we launched from $y = 0$ , the final vertical displacement is also zero.

Step 1: Find the time of flight ( $T$ )
We use the vertical position equation and set $y (t) = 0$ and solve for $t = T$ :
$y (T) = (v_{0} sin θ) T - \frac{1}{2} g T^{2} = 0$
This is a quadratic equation for $T$ . We can factor out $T$ :
$T (v_{0} sin θ - \frac{1}{2} g T) = 0$
This equation has two solutions:

$T = 0$ , which corresponds to the initial launch moment.
$v_{0} sin θ - \frac{1}{2} g T = 0$ , which gives $T = \frac{2 v _{0} s i n θ}{g}$ .

The second, non-trivial solution is our time of flight:
$T = \frac{2 v _{0} s i n θ}{g}$
Notice that $T = 2 \times t_{p e ak}$ . This mathematical result reflects the symmetry of the parabolic path: the time taken to rise to the peak is equal to the time taken to fall back to the original height.

Step 2: Calculate the horizontal range ( $R$ )
The range is the horizontal position $x$ at time $t = T$ . We use the horizontal position equation:
$R = x (T) = (v_{0} cos θ) T$
Substitute the expression for $T$ :
$R = (v_{0} cos θ) (\frac{2 v _{0} s i n θ}{g})$
$R = \frac{2 v _{0}^{2} s i n θ c o s θ}{g}$
Using the trigonometric double-angle identity $2 sin θ cos θ = sin (2 θ)$ , we arrive at the standard formula for the range:
$R = \frac{v _{0}^{2} s i n ( 2 θ )}{g}$

The video below provides an excellent visual walkthrough of the derivations we have just completed.

Projectile Motion: Finding the Maximum Height and the Range

To see these derivations presented dynamically, please watch this video from the Physics Ninja channel. It clearly explains the logic for finding the time to peak, maximum height, total flight time, and range.

Watch from 11:37 to 19:12. The first part (11:37 - 16:16) covers the derivation for maximum height. The second part (16:16 - 19:12) covers the time of flight and horizontal range.

3. Alternative Proofs and Analysis

Since you prefer rigorous and varied proofs, let's explore an alternative method that builds directly on the trajectory equation $y (x)$ from our last lesson. This approach uses calculus on the trajectory equation itself.

Integration and projectile motion (Sect. 13.2)

The following document from Michigan State University provides a concise, theorem-proof style derivation for these same quantities, starting from the trajectory equation. This offers a different and complementary mathematical perspective.

Please read the section titled 'Range, Height, Flight Time'. Focus on the 'Proof' subsections that start from the equation y(x). Notice how the range is found by setting y(xr) = 0, and the height is found by finding the x-coordinate where y'(x) = 0.

The proof for maximum height in this document leverages a key concept from calculus: the maximum of a function occurs where its derivative is zero. For the trajectory $y (x)$ , the maximum height is at the $x$ -coordinate where $d y / d x = 0$ . This reinforces the connection between the physical condition ( $v_{y} = 0$ ) and the geometric properties of the trajectory.

Analysis of the Range Formula

The formula for range, $R = \frac{v _{0}^{2} s i n ( 2 θ )}{g}$ , reveals some interesting physics. For a fixed initial speed $v_{0}$ , the range depends only on the launch angle $θ$ .

Maximum Range: The range is maximized when $sin (2 θ)$ is maximized. The maximum value of the sine function is 1, which occurs when its argument is $9 0^{\circ}$ .
$2 θ = 9 0^{\circ} ⟹ θ = 4 5^{\circ}$
Thus, neglecting air resistance, a 45° launch angle achieves the maximum possible horizontal range.

Test your understanding!

Prove that the horizontal range is the same for two launch angles that are complementary, i.e., $θ_{1}$ and $θ_{2} = 9 0^{\circ} - θ_{1}$ . (Assume $0 < θ_{1} < 9 0^{\circ}$ ).

Show answer

Let's write the range for each angle.

For the first angle, $θ_{1}$ :
$R_{1} = \frac{v _{0}^{2} s i n ( 2 θ _{1} )}{g}$

For the second angle, $θ_{2} = 9 0^{\circ} - θ_{1}$ :
$R_{2} = \frac{v _{0}^{2} s i n ( 2 θ _{2} )}{g} = \frac{v _{0}^{2} s i n ( 2 ( 9 0 ^{\circ} - θ _{1} ))}{g}$
$R_{2} = \frac{v _{0}^{2} s i n ( 18 0 ^{\circ} - 2 θ _{1} )}{g}$
Using the trigonometric identity $sin (18 0^{\circ} - α) = sin (α)$ , we get:
$R_{2} = \frac{v _{0}^{2} s i n ( 2 θ _{1} )}{g}$
Therefore, $R_{1} = R_{2}$ . This proves that complementary launch angles yield the same horizontal range. For example, a launch at 30° and a launch at 60° will land at the same spot, although their maximum heights and times of flight will be different.

Conclusion

In this lesson, we have systematically derived the three most important descriptive quantities for projectile motion from first principles.

Key Takeaways:

The maximum height is found using the condition that vertical velocity is zero at the peak:
$H = \frac{( v _{0} s i n θ ) ^{2}}{2 g}$
The time of flight (for a level ground scenario) is found using the condition that the final vertical displacement is zero:
$T = \frac{2 v _{0} s i n θ}{g}$
The horizontal range is the horizontal distance covered during the time of flight:
$R = \frac{v _{0}^{2} s i n ( 2 θ )}{g}$
These formulas are interconnected, and their derivations highlight the symmetrical nature of the trajectory.
The maximum range for a given initial speed is achieved at a launch angle of 45°.

Preview of the Next Lesson:
The formulas we derived today rely on the crucial assumption that the launch and landing points are at the same vertical height. What happens if this is not the case? In our next lesson, we will tackle projectile motion on an inclined plane. This will require us to be more creative with our choice of coordinate systems and will demonstrate that a deep understanding of the principles of motion is more powerful than memorizing formulas.

Recommended resources

1. Maximum Height (H)

Physical Condition and Derivation

Alternative Derivation

2. Time of Flight (T) and Horizontal Range (R)

Physical Condition and Derivation

3. Alternative Proofs and Analysis

Analysis of the Range Formula

Conclusion

1. Maximum Height ( $H$ )

2. Time of Flight ( $T$ ) and Horizontal Range ( $R$ )