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PROJECTILE MOTION

Understanding Horizontal & Vertical Components of Velocity

Learn how gravity affects only the vertical component of a projectile's velocity, why the trajectory is a parabola, and how angled throws change the path.

The Scenario

A ball is thrown horizontally from the top of a hill.

As it leaves the thrower's hand, it has:

• Initial horizontal velocity (Vx)
• Zero initial vertical velocity (Vy = 0)

Gravity pulls it downward, creating a curved parabolic path to the ground.

Horizontal Component (Vx)

Key Concept
Gravity acts DOWNWARD (vertical). The horizontal direction is perpendicular (90°) to gravity.
Therefore, gravity has NO effect on horizontal velocity. Vx remains constant.
Force & Direction Diagram
Horizontal Motion Equations
Acceleration (ax) = 0
(No horizontal force)
Vx = Vx0 (constant)
Horizontal Distance:
x = Vx0 × t

Vertical Component (Vy)

Key Concept
Gravity acts in the SAME direction as the vertical component (both downward).
So gravity continuously accelerates the ball downward. Vy increases over time!
Vertical Acceleration
t=0
0.0
t=1s
9.8
t=2s
19.6
t=3s
29.4
Vy grows each second ↓
Vertical Motion Equations
Acceleration (ay) = g = 9.8 m/s²
(Gravity pulls downward)
Vy = g × t
(since initial Vy = 0)
Height fallen:
h = ½ × g × t²

Think First!

Answer the question below before we reveal the concept
Quick Question
A ball is thrown horizontally from the top of a hill. What will be the shape of its path as it falls to the ground?
A
Straight line (angled downward)
B
Parabola
C
Depends on the velocity at which the ball is thrown
D
Depends on the height of the hill

Interactive Simulation

Adjust initial horizontal velocity and observe the trajectory
Controls
5101520 m/s
Vx0 = 10 m/s
2035506580 m
h = 45 m
Live Readouts
Time: 0.0 s
Vx: 10 m/s (constant)
Vy: 0.0 m/s
Range: — m
y = 0.049 × x² (parabola)
Simulation Canvas

Test Your Understanding

Question 1
A ball is thrown horizontally from a hill of height 80 m with an initial horizontal velocity of 15 m/s. What is the horizontal distance from the base of the hill where the ball lands?
(Take g = 10 m/s²)
A
45 meters
B
60 meters
C
75 meters
D
120 meters
Hint: First find time using h = ½ g t², then use x = Vx0 × t

Result — Question 1

SOLUTION BREAKDOWN

Given: h = 80 m,  Vx0 = 15 m/s,  g = 10 m/s²
Step 1: Find time of flight
h = ½ × g × t²  →  80 = ½ × 10 × t²  →  t² = 16  →  t = 4 seconds
Step 2: Find horizontal distance
x = Vx0 × t = 15 × 4 = 60 meters
Gravity doesn't affect Vx, so it stays 15 m/s the entire flight!

PART 2: THROW AT AN ANGLE

What happens when the ball is thrown at an angle — upward (+θ) or downward (−θ)?

Throwing at an Angle (θ)

Key Concept
When thrown at angle θ with speed V0, the velocity has BOTH components from the start:
Vx0 = V0 × cos(θ)     Vy0 = V0 × sin(θ)
Positive Angle (+θ): Ball Thrown UP
Vy0 is UPWARD (positive)
Ball first rises, slows down, stops momentarily, then falls
Gravity opposes upward motion:
Vy = V0 sin(θ) – g × t
Longer flight time → Greater horizontal range
Path: still a parabola, but inverted ∩ shape that goes up first then curves down below hilltop level
Negative Angle (−θ): Ball Thrown DOWN
Vy0 is DOWNWARD (negative)
Ball immediately goes down and accelerates faster
Gravity adds to downward motion:
Vy = V0 sin(θ) + g × t  (both downward)
Shorter flight time → Smaller horizontal range
Path: steeper parabola curving sharply downward

Angled Throw: Complete Equations

Initial Velocity Components
Initial velocity V0 at angle θ:   Vx0 = V0 cosθ     Vy0 = V0 sinθ
Horizontal (unchanged!)
ax = 0  (still no horizontal force)
Vx = V0 cosθ (constant)
x(t) = V0 cosθ × t
Same principle: gravity is perpendicular, so Vx is unaffected!
Vertical (now has initial Vy!)
ay = g  (gravity still acts)
Vy(t) = V0 sinθ + g × t
y(t) = V0 sinθ × t + ½ g t²
+θ → sinθ > 0 (upward)  |  −θ → sinθ < 0 (downward)
Path Equation (still a parabola!)
y = x × tanθ + (g / 2V0²cos²θ) × x²
This is y = ax + bx² — still quadratic in x, so the path is ALWAYS a parabola!

How Angle Changes the Path

Trajectory Comparison (V0 = 15 m/s, h = 44.1 m)
All three paths are parabolas. Throwing upward (+θ) gives the ball more time in the air → greatest range. Throwing downward (−θ) reduces flight time → shortest range. In all cases, Vx remains constant — only Vy is affected by gravity.

Simulation: Angled Throw

Change speed and angle to see how the trajectory changes
Controls
510152025 m/s
V0 = 20 m/s
−60°−30°0°+30°+60°
θ = +30° (upward)
2035506580 m
h = 45 m
Computed Values
Vx0 = 17.3 m/s
Vy0 = 10.0 m/s ↑
Time of flight: 4.2 s
Range: 72.7 m
y = −0.577 × x + 0.0163 × x² (parabola)
Simulation Canvas (θ = +30°)

Test Your Understanding — Angled Throw

Question 2
A ball is thrown from a 45 m high hill at 20 m/s at an angle of +30° above the horizontal. What is the horizontal distance from the base of the hill where the ball lands?
(Take g = 10 m/s².  cos30° = 0.866,  sin30° = 0.5)
A
52 meters
B
69.3 meters
C
86.6 meters
D
100 meters
Hint: Decompose V0 into components, find t from the vertical equation, then use x = Vx0 × t

Result — Question 2

SOLUTION BREAKDOWN

Given: V0 = 20 m/s,  θ = +30°,  h = 45 m,  g = 10 m/s²
Step 1: Decompose velocity
Vx0 = 20 × cos30° = 20 × 0.866 = 17.32 m/s
Vy0 = 20 × sin30° = 20 × 0.5 = 10 m/s (upward)
Step 2: Find time of flight (taking downward as +ve)
h = −Vy0 × t + ½ g t²
45 = −10t + 5t²  →  5t² − 10t − 45 = 0  →  t² − 2t − 9 = 0
t = (2 + √(4+36)) / 2 = (2 + √40) / 2 ≈ 4 seconds
Step 3: Horizontal distance
x = Vx0 × t = 17.32 × 4 ≈ 69.3 meters
The upward throw gave the ball more time in air (4s vs 3s for horizontal), resulting in a greater range!