MechanicsR = v₀² sin2θ / g

Projectile Motion Simulator

Simulate projectile motion with adjustable launch angle and speed. Visualize trajectory, calculate range, max height, and time of flight interactively.

— vₓ— vᵧ— trail

Parameters

Launch Speed v₀

m/s

Launch Angle θ

Gravity g

m/s²

Predicted

Range R91.84 m

Max Height H22.96 m

Flight Time T4.33 s

Time0.00 s

x0.00 m

y0.00 m

vₓ0.00 m/s

vᵧ0.00 m/s

Speed0.00 m/s

About Projectile Motion

Projectile motion describes the curved path of an object launched into the air and moving under gravity alone (ignoring air resistance). The key insight is that horizontal and vertical motions are completely independent of each other.

ℹAny object in projectile motion follows a parabolic path. The horizontal component has constant velocity; the vertical component has constant acceleration (−g).

Two Independent Components

Horizontal (x): Constant velocity — no acceleration. vₓ = v₀ cos θ remains unchanged throughout.
Vertical (y): Uniform downward acceleration. vᵧ starts at v₀ sin θ and decreases by g each second.

Key Variables

Symbol	Name	Unit	Description
v₀	Initial Speed	m/s	Speed at the moment of launch
θ	Launch Angle	°	Angle above the horizontal
g	Gravitational Acceleration	m/s²	9.8 m/s² on Earth; 1.6 m/s² on Moon
R	Range	m	Horizontal distance from launch to landing
H	Maximum Height	m	Highest point above the launch level
T	Time of Flight	s	Total time the projectile is in the air
t	Elapsed Time	s	Time since launch at any instant
x(t)	Horizontal Position	m	Horizontal distance at time t
y(t)	Vertical Position	m	Height at time t

Worked Example 1 — Range at 45°

Ball launched at v₀ = 30 m/s, θ = 45°, g = 9.8 m/s²:

R = \frac{v _{0}^{2} sin 2 θ}{g} = \frac{3 0 ^{2} \times sin 90°}{9.8} \approx 91.8 m

H = \frac{v _{0}^{2} sin ^{2} θ}{2 g} = \frac{3 0 ^{2} \times 0.5}{2 \times 9.8} \approx 22.9 m

T = \frac{2 v _{0} sin θ}{g} = \frac{2 \times 30 \times sin 45°}{9.8} \approx 4.33 s

Worked Example 2 — Maximum Range

Range is maximised when sin(2θ) = 1, i.e., 2θ = 90°, so θ = 45°. Complementary angles (e.g. 30° and 60°) always give the same range because sin(2 × 30°) = sin(2 × 60°).

💡Try g = 1.6 m/s² (Moon gravity) in the simulator and compare the range to Earth. On the Moon, the same ball travels over 6× farther!

Tips for Using the Simulator

The dashed blue arc is the predicted full trajectory — it updates instantly as you move sliders.
The orange ball with the green trail shows the simulated path during animation.
Blue arrow = vₓ (constant horizontal velocity). Pink arrow = vᵧ (decreases to zero at peak, then reverses).
The Predicted panel on the right shows R, H, and T before you even launch.
After landing, the ball rests at the exact calculated range — compare with the predicted value.

Key Formulas

Formula	Description	Notes
x(t) = v₀ cosθ · t	Horizontal position	Constant velocity — no x-acceleration
y(t) = v₀ sinθ · t − ½gt²	Vertical position	Parabolic — rises then falls
R = v₀² sin2θ / g	Range	Maximum at θ = 45°
H = v₀² sin²θ / (2g)	Maximum height	Reached when vᵧ = 0
T = 2v₀ sinθ / g	Time of flight	Twice the time to peak
vₓ = v₀ cosθ	Horizontal velocity	Constant throughout flight
vᵧ = v₀ sinθ − gt	Vertical velocity	Zero at max height

Equation of Motion (Horizontal)

x (t) = v_{0} cos θ \cdot t

Equation of Motion (Vertical)

y (t) = v_{0} sin θ \cdot t - \frac{1}{2} g t^{2}

Range, Max Height, Time of Flight

R = \frac{v _{0}^{2} sin 2 θ}{g} H = \frac{v _{0}^{2} sin ^{2} θ}{2 g} T = \frac{2 v _{0} sin θ}{g}

Velocity Components at Any Time t

v_{x} = v_{0} cos θ v_{y} (t) = v_{0} sin θ - g t

Resultant Speed

∣ v ∣ = v_{x}^{2} + v_{y}^{2}

ℹAt maximum height, vᵧ = 0 and the speed equals vₓ = v₀ cosθ — the minimum speed during the flight.

Frequently Asked Questions

What angle gives maximum range in projectile motion?

45° gives the maximum range on flat ground (in the absence of air resistance), because sin(2 × 45°) = sin(90°) = 1, its maximum value.

Does a heavier object land sooner than a lighter one?

No. In the absence of air resistance, all objects fall at the same rate regardless of mass. Both would follow identical trajectories if launched with the same initial speed and angle.

Why is the path parabolic?

Because x increases linearly with time while y is a quadratic function of time. Eliminating t from x(t) and y(t) gives y as a quadratic function of x — the equation of a parabola.

Why do 30° and 60° give the same range?

Because sin(2 × 30°) = sin(60°) and sin(2 × 60°) = sin(120°) = sin(60°). Complementary launch angles always produce the same range.

What is the speed at maximum height?

At maximum height vᵧ = 0, so the speed equals the horizontal component: v₀ cosθ. This is the minimum speed during the entire flight.