Q.A projectile is launched with an initial speed of 15 m/s, once at $30^\circ$ and once at $60^\circ$ to the horizontal (take $g = 10\ \text{m/s}^2$). Show, by calculating both ranges, that the two ranges are equal, and explain why this happens for any pair of complementary launch angles.

Projectile Motion — From Intuition to Precision

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it rises, slows down, then curves downward and falls. That curved path is a projectile's trajectory. The ball is a projectile: any object that is launched into the air and then moves only under the influence of gravity (and air resistance, which we ignore for now).

The key intuition: once the ball leaves your hand, the only force acting on it is gravity pulling it straight down. There is no forward force after release. The ball keeps moving forward because of inertia — it wants to keep going in a straight line at constant speed. But gravity keeps pulling it down, so the forward motion and downward acceleration combine to produce a curved path.

The Precise Statement

We break the motion into two independent components:

• Horizontal motion: No acceleration ($a_x = 0$). So horizontal velocity $v_x$ is constant.
• Vertical motion: Constant downward acceleration ($a_y = -g$). So vertical velocity $v_y$ changes linearly with time.

The independence of these components is the central idea — what happens vertically does not affect what happens horizontally, and vice versa.

The Equations (for a projectile launched with initial speed $u$ at angle $\theta$ above horizontal)

First, resolve the initial velocity:

Horizontal motion (constant velocity):

Vertical motion (constant acceleration $-g$):

Key Results You Must Know

The Trajectory Equation (Path Shape)

Eliminate $t$ from the $x$ and $y$ equations to get $y$ as a function of $x$:

This is a parabola — the signature shape of projectile motion.

Common Mistake to Avoid …