Range of a Projectile

What this formula says

A projectile is launched from the ground with speed $u$ at angle $\theta$. It lands back on the same level. The horizontal distance between launch and landing is the range:

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

Setting up the problem

In projectile motion, horizontal and vertical motions are independent.

• Horizontal: no acceleration, constant velocity $u_x = u\cos\theta$
• Vertical: constant downward acceleration $g$, initial velocity $u_y = u\sin\theta$

Range is a horizontal quantity — it equals the horizontal velocity multiplied by the total time of flight.

$R = u_x \times T = u\cos\theta \times T$

We already know the time of flight:

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

Derivation

Substitute $T$ into the range equation:

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

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

Now use the double angle identity: $2\sin\theta\cos\theta = \sin 2\theta$

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

Why the double angle identity appears

The identity $\sin 2\theta = 2\sin\theta\cos\theta$ is a standard trigonometric result. It appears here naturally because the range depends on both $\sin\theta$ (from the vertical motion, via $T$) and $\cos\theta$ (from the horizontal motion). Their product $\sin\theta\cos\theta$ is exactly half of $\sin 2\theta$.

Maximum range

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

$\sin 2\theta$ is maximum when $2\theta = 90°$, i.e. when $\theta = 45°$.

At $\theta = 45°$: $\sin 90° = 1$, so:

$R_{\text{max}} = \frac{u^2}{g}$

Complementary angles give the same range

If $\theta_1 + \theta_2 = 90°$, then $\theta_2 = 90° - \theta_1$, and:

$\sin 2\theta_2 = \sin(180° - 2\theta_1) = \sin 2\theta_1$

So both angles give the same range. For example, $30°$ and $60°$ give identical ranges. So do $20°$ and $70°$, or $15°$ and $75°$.

How range varies with angle

Assumptions

• Launch and landing at the same height
• No air resistance
• $g$ constant throughout