Projectiles (Edexcel A-Level Mathematics): Revision Notes

2.6.4 Deriving Projectile Formulae

Projectile motion involves objects moving in a curved path under the influence of gravity, typically neglecting air resistance. The motion can be analysed by breaking it into two components: horizontal and vertical. The horizontal motion has constant velocity, and the vertical motion has constant acceleration due to gravity.

1. Basic Assumptions

2. Splitting the Initial Velocity into Components

If a projectile is launched with an initial velocity $u$ at an angle $\theta$ to the horizontal:

• Horizontal Component: $u_x = u \cos \theta$
• Vertical Component: $u_y = u \sin \theta$

3. Horizontal Motion

For the horizontal motion:

• Acceleration: $a_x = 0$ (no horizontal acceleration)
• Velocity: $v_x = u_x = u \cos \theta$ (remains constant)
• Displacement: $x = u_x t = (u \cos \theta) t$

4. Vertical Motion

For the vertical motion:

• Acceleration: $a_y = -g$ (due to gravity)
• Initial Velocity: $u_y = u \sin \theta$
• Final Velocity: $v_y = u \sin \theta - gt$
• Displacement: $y = u \sin \theta \cdot t - \frac{1}{2} g t^2$

5. Deriving Key Projectile Formulae

a. Time of Flight ($T$)

The time of flight is the total time the projectile remains in the air. To find it, consider the time it takes for the projectile to return to the same vertical level from which it was launched (assuming it lands at the same height):

At maximum height, $v_y = 0$:

$v_y = u \sin \theta - gt = 0$

$t = \frac{u \sin \theta}{g}$

This is the time to reach maximum height. The total time of flight is twice this time:

$T = \frac{2u \sin \theta}{g}$

b. Maximum Height ($H$)

The maximum height $H$ is the highest point the projectile reaches. Using the equation:

$v_y^2 = u_y^2 - 2gH$

At maximum height, $v_y = 0$:

$0 = (u \sin \theta)^2 - 2gH$

$H = \frac{(u \sin \theta)^2}{2g}$

c. Range ($R$)

The range $R$ is the horizontal distance the projectile covers. To find it, use the horizontal displacement formula $x = u_x t$ , and substitute the time of flight $T$ :

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

Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$, the range can be simplified to:

$R = \frac{u^2 \sin 2\theta}{g}$

6. Summary of Key Formulae

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