Speed of Projectile at Any Time

What this formula gives

At any instant $t$ during flight, the projectile has:

• Horizontal velocity: $v_x = u\cos\theta$
• Vertical velocity: $v_y = u\sin\theta - gt$

These two components are perpendicular to each other. The actual speed — the magnitude of the velocity vector — is found by the Pythagorean theorem:

$|\vec{v}| = \sqrt{v_x^2 + v_y^2}$

$\boxed{|\vec{v}| = \sqrt{u^2\cos^2\theta + (u\sin\theta - gt)^2}}$

Derivation

The velocity vector at time $t$ has components $v_x$ and $v_y$. Since they are perpendicular:

$|\vec{v}|^2 = v_x^2 + v_y^2$

Substitute $v_x = u\cos\theta$ and $v_y = u\sin\theta - gt$:

$|\vec{v}|^2 = (u\cos\theta)^2 + (u\sin\theta - gt)^2$

$|\vec{v}| = \sqrt{u^2\cos^2\theta + (u\sin\theta - gt)^2}$

Expanding the expression

Expand $(u\sin\theta - gt)^2 = u^2\sin^2\theta - 2ugt\sin\theta + g^2t^2$:

$|\vec{v}|^2 = u^2\cos^2\theta + u^2\sin^2\theta - 2ugt\sin\theta + g^2t^2$

Since $\cos^2\theta + \sin^2\theta = 1$:

$|\vec{v}|^2 = u^2 - 2ugt\sin\theta + g^2t^2$

$\boxed{|\vec{v}| = \sqrt{u^2 - 2ugt\sin\theta + g^2t^2}}$

This expanded form is often more useful for calculations.

How speed varies during flight

At launch ($t = 0$): $|\vec{v}| = u$ — full launch speed.

As the projectile rises, $v_y$ decreases. Since $v_x$ stays constant, the speed decreases.

At the top: $v_y = 0$, so speed = $|v_x| = u\cos\theta$ — this is the minimum speed during flight.

As the projectile descends, $v_y$ increases in magnitude (downward). Speed increases back toward $u$.

At landing: By symmetry, speed = $u$ again — same as launch speed.

Special case: $\theta = 0$ (horizontal projection)

$|\vec{v}| = \sqrt{u^2 + g^2t^2}$

Speed starts at $u$ and increases throughout — there is no upward component to first reduce the speed.