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Projectile Motion Calculator

Enter initial velocity, launch angle, and height to compute the trajectory. Drag the time slider to step through the flight path and see position, velocity, and direction at any moment. All calculations run in your browser.

Parameters

m/s
°
m
m/s²

Results

Time of Flight
2.8832 s
Max Height
10.1937 m
Range
40.7747 m
Time to Max Height
1.4416 s
Impact Velocity
20 m/s
Impact Angle
45°

Trajectory

0 s
0 s2.8832 s

Values at t = 0 s

x
0 m
y
0 m
vₓ
14.1421 m/s
vᵧ
14.1421 m/s
Speed
20 m/s

Step-by-step breakdown

θ=45∘=0.7854 rad\theta = 45^\circ = 0.7854 \text{ rad}
v0x=v0cos⁡θ=20×0.7071=14.1421 m/sv_{0x} = v_0 \cos\theta = 20 \times 0.7071 = 14.1421 \text{ m/s}
v0y=v0sin⁡θ=20×0.7071=14.1421 m/sv_{0y} = v_0 \sin\theta = 20 \times 0.7071 = 14.1421 \text{ m/s}
tmax=v0yg=14.14219.81=1.4416 st_{\text{max}} = \frac{v_{0y}}{g} = \frac{14.1421}{9.81} = 1.4416 \text{ s}
H=h0+v0y22g=0+14.142122×9.81=10.1937 mH = h_0 + \frac{v_{0y}^2}{2g} = 0 + \frac{14.1421^2}{2 \times 9.81} = 10.1937 \text{ m}
T=v0y+v0y2+2gh0g=14.1421+2009.81=2.8832 sT = \frac{v_{0y} + \sqrt{v_{0y}^2 + 2gh_0}}{g} = \frac{14.1421 + \sqrt{200}}{9.81} = 2.8832 \text{ s}
R=v0x×T=14.1421×2.8832=40.7747 mR = v_{0x} \times T = 14.1421 \times 2.8832 = 40.7747 \text{ m}
vimpact=v0x2+vy,T2=20 m/sv_{\text{impact}} = \sqrt{v_{0x}^2 + v_{y,T}^2} = 20 \text{ m/s}
θimpact=45∘\theta_{\text{impact}} = 45^\circ

Reference Guide

Projectile Motion Basics

A projectile moves in two independent directions. Horizontal velocity stays constant because no horizontal force acts on it (ignoring air resistance). Vertical motion follows free fall under gravity.

The combination of constant horizontal speed and accelerating vertical motion produces a parabolic path. This separation of motion into independent axes is one of the most useful ideas in physics.

Horizontal
x(t)=v0cos⁡θ⋅tx(t) = v_0 \cos\theta \cdot t
Vertical
y(t)=h0+v0sin⁡θ⋅t−12gt2y(t) = h_0 + v_0 \sin\theta \cdot t - \tfrac{1}{2}g t^2

Key Equations

Position
x=v0cos⁡θ⋅t,y=h0+v0sin⁡θ⋅t−12gt2x = v_0 \cos\theta \cdot t, \quad y = h_0 + v_0 \sin\theta \cdot t - \tfrac{1}{2}g t^2
Velocity
vx=v0cos⁡θ,vy=v0sin⁡θ−gtv_x = v_0 \cos\theta, \quad v_y = v_0 \sin\theta - g t
Time of flight Solve y(T)=0y(T) = 0 for the positive root
T=v0sin⁡θ+(v0sin⁡θ)2+2gh0gT = \frac{v_0 \sin\theta + \sqrt{(v_0 \sin\theta)^2 + 2 g h_0}}{g}

Maximum Height and Range

The projectile reaches maximum height when vertical velocity becomes zero. The range is how far it travels horizontally before landing.

Max height
H=h0+(v0sin⁡θ)22gH = h_0 + \frac{(v_0 \sin\theta)^2}{2g}
Range
R=v0cos⁡θ⋅TR = v_0 \cos\theta \cdot T

On flat ground (h0=0h_0 = 0), a 45-degree launch angle gives the maximum range. When launching from a height, the optimal angle is less than 45 degrees.

Velocity and Energy

Speed changes throughout the flight because the vertical component changes while horizontal remains constant. The projectile moves slowest at the apex, where only horizontal velocity remains.

Speed at any time
∣v∣=vx2+vy2|v| = \sqrt{v_x^2 + v_y^2}

By energy conservation, a projectile launched from ground level returns with the same speed it started with. When launched from a height, it impacts faster because gravity adds energy during the extra fall.

Energy relation
12mv02+mgh0=12mvimpact2\tfrac{1}{2}m v_0^2 + m g h_0 = \tfrac{1}{2}m v_{\text{impact}}^2