Projectile Velocity Calculator

A projectile has one velocity, but you almost never work with it directly. Split it into components that behave independently:

$v_x = v_0\cos\theta \quad (\text{constant}), \qquad v_y = v_0\sin\theta - gt$

• $v_0$ — launch speed, m/s; $\theta$ — launch angle above the horizontal
• $g = 9.81$ m/s²; $t$ — time since launch, s

Horizontal velocity never changes without air resistance, because nothing pushes horizontally. Vertical velocity falls at $9.81$ m/s every second, passes through zero at the apex, and is negative on the way down.

Recombine when the question asks for speed or direction:

$v = \sqrt{v_x^2 + v_y^2}, \qquad \phi = \tan^{-1}\!\left(\frac{v_y}{v_x}\right)$

The assumption people forget: $\cos$ goes with horizontal and $\sin$ with vertical only when $\theta$ is measured from the horizontal. If a problem quotes the angle from the vertical, the two swap.

Most exam questions give you an outcome and want the launch. The four kinematic equations, applied to the vertical axis with $a = -g$, cover every case:

$v = v_0 + at, \qquad s = v_0t + \tfrac12 at^2, \qquad v^2 = v_0^2 + 2as$

with $s$ in metres, $v$ in m/s and $a$ in m/s². Two shortcuts save time:

• From the range on level ground: $R = \dfrac{v_0^2 \sin 2\theta}{g}$, so $v_0 = \sqrt{\dfrac{Rg}{\sin 2\theta}}$
• From the maximum height: $H = \dfrac{v_0^2\sin^2\theta}{2g}$, so $v_0\sin\theta = \sqrt{2gH}$

On an inclined plane the same equations apply along the slope, with the acceleration component $a = g\sin\theta$ down the incline (frictionless) and $g\cos\theta$ pressing into it.

When it applies: constant acceleration only. The moment air drag matters these equations stop being exact.

• Using $g$ on the horizontal axis — horizontal acceleration is zero, so $v_x$ is the same at launch, apex and landing.
• Swapping sine and cosine — check against a limiting case: at $\theta = 0$ the launch is purely horizontal, and only $\cos 0 = 1$ gives that.
• Treating speed at the apex as zero — only the vertical component vanishes there. The projectile is still moving at $v_x$.
• Adding components arithmetically — combine them with Pythagoras, not by addition.
• Forgetting the sign of $v_y$ on descent — it is negative, and dropping the minus sign hides the direction of travel.
• Using $\sin 2\theta$ off level ground — the range formula assumes launch and landing at the same height.