Q.If one object is dropped vertically downward and another object is thrown horizontally from the same height, then the ratio of the vertical distance covered by both objects at any instant $t$ is

Projectile Motion Under Gravity

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it curves upward, then arcs downward. That curve is a parabola, and the motion is called projectile motion.

The key insight: once the ball leaves your hand, the only force acting on it (ignoring air resistance) is gravity pulling it straight down. There is no force pushing it sideways or upward after release. That single downward force is what creates the beautiful curved path.

The Core Idea

A projectile is any object that is thrown, launched, or otherwise projected into the air and then moves under the influence of gravity alone. The motion has two independent parts happening simultaneously:

1. Horizontal motion: constant speed (no horizontal force)
2. Vertical motion: constant downward acceleration $g \approx 9.8 \, \text{m/s}^2$

These two motions are completely independent — they don't affect each other. This is the most important thing to understand.

Breaking It Down Mathematically

Let's set up coordinates: $x$ is horizontal, $y$ is vertical (positive upward). The launch point is at $(0,0)$ with initial speed $u$ at angle $\theta$ above horizontal.

Initial velocity components:

Horizontal motion (no acceleration):

Vertical motion (constant downward acceleration $g$):

The minus sign is because gravity pulls downward, opposite to our positive $y$ direction.

The Path Is a Parabola

Eliminate $t$ between the $x$ and $y$ equations. From $x = u_x t$, we get $t = \frac{x}{u \cos \theta}$. Substitute into the $y$ equation:

This is of the form $y = ax - bx^2$, which is a parabola opening downward. That's why every projectile under gravity follows a parabolic path.

Key Quantities You'll Need

Time of Flight ($T$)

The total time the projectile stays in the air. Set $y = 0$ (returns to launch height):

Factor $T$: $T (u \sin \theta - \frac{1}{2} g T) = 0$

The non-zero solution:

Maximum Height ($H$)

The highest point occurs when vertical velocity becomes zero: $v_y = u \sin \theta - g t = 0$, so $t = \frac{u \sin \theta}{g}$.

Plug into $y$ equation:

Range ($R$)

Horizontal distance traveled when it returns to launch height. Use $x = u_x T$:

Using $\sin 2\theta = 2 \sin \theta \cos \theta$:

Common Mistakes to Avoid