Projectile Motion

You’re a passenger in a car going 60 km/h and you toss a ball straight up. To you, the ball goes straight up and comes straight back down into your hand. To someone standing on the sidewalk watching you go by, the ball traces a beautiful arc through the air — a parabola.

Both observers are right. The ball is doing two motions at once: a horizontal one (the car’s speed, which the ball still has) and a vertical one (your toss + gravity). Those two motions don’t interfere with each other. They just happen at the same time.

That single idea — horizontal and vertical motion are independent — is the master key to every projectile problem on the MCAT. Master it and the rest is just bookkeeping.

Use the simulation below to explore how launch angle and initial speed change the trajectory. Watch the horizontal velocity (blue) stay rock-steady while the vertical velocity (red) shrinks, hits zero, then grows in the other direction.

The Two Independent Motions

Every projectile problem splits into two parts:

| Direction | Acceleration | Velocity |
|-----------|-------------|----------|
| Horizontal (x) | 0 (no horizontal force) | Constant: $v_x = v_0 \cos\theta$ |
| Vertical (y) | −g (gravity pulls down) | Changes: $v_y = v_0 \sin\theta - gt$ |

The horizontal direction has no acceleration (no force pushes the ball sideways once it’s left your hand), so the horizontal velocity never changes. The vertical direction has constant downward acceleration $g$ — exactly the free fall problem from §1.5.

Solving Projectile Problems: Step by Step

1. Resolve the initial velocity into horizontal ($v_0 \cos\theta$) and vertical ($v_0 \sin\theta$) components.
2. Use the vertical equations to find time of flight, max height, or vertical position.
3. Plug that time into the horizontal equation to find range or horizontal position.

That’s it. Two motions, one shared variable (time), three steps.

Key Projectile Relationships

For a projectile launched from and landing at the same height (level ground):

Launch Angle Effects

The launch angle dramatically changes the trajectory. Soccer players, kickers, and quarterbacks all unconsciously do this math:

| Angle | Range | Height | Notes |
|-------|-------|--------|-------|
| 45° | Maximum | Moderate | Optimizes range because $\sin(2 \times 45°) = \sin 90° = 1$ |
| 30° | Same as 60° | Lower than 60° | Complementary angles share range |
| 60° | Same as 30° | Higher than 30° | Complementary angles share range |
| 0° | — | 0 | Horizontal launch (rolled off a table) |
| 90° | 0 | Maximum | Straight up and down |

Special Case: Horizontal Launch

When something rolls off a table or flies off a cliff horizontally, the initial vertical velocity is zero. The object then falls like any dropped object vertically, while keeping its horizontal speed.

• Time to hit the ground depends only on the height: $h = \tfrac{1}{2}gt^2$, so $t = \sqrt{2h/g}$.
• Horizontal distance = (horizontal speed) × (time). Faster sideways speed → lands farther away. But the fall time itself doesn’t change — it’s set entirely by the height.

Famous demo: shoot a bullet horizontally and drop a second bullet from the same height at the same instant. They both hit the ground at the same time. The shot bullet just lands much farther away.