Horizontal Projection — Velocity at Any Time

The situation

A body projected horizontally with speed $u$ has at time $t$:

• Horizontal velocity: $v_x = u$ (constant, no horizontal acceleration)
• Vertical velocity: $v_y = gt$ (starts at zero, increases as it falls)

The speed is the magnitude of the velocity vector:

$|\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + (gt)^2}$

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

Derivation

Horizontal: $v_x = u + 0 \cdot t = u$

Vertical (taking downward positive, starting from rest): $v_y = 0 + gt = gt$

By Pythagoras:

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

How speed changes with time

At $t = 0$: speed $= u$ (purely horizontal)

As $t$ increases: $v_y = gt$ grows, so speed increases continuously.

At landing ($t = T = \sqrt{\frac{2h}{g}}$):

$|\vec{v}| = \sqrt{u^2 + g^2 \cdot \frac{2h}{g}} = \sqrt{u^2 + 2gh}$

Unlike a standard projectile launched at an angle, the speed of a horizontally projected body never decreases — it starts at $u$ and grows throughout the flight.

Angle with the horizontal at time $t$

$\tan\phi = \frac{v_y}{v_x} = \frac{gt}{u}$

$\phi = \tan^{-1}\left(\frac{gt}{u}\right)$

At launch: $\phi = 0°$ (purely horizontal). As the body falls, $\phi$ increases — the velocity vector tilts more and more downward.

At landing:

$\tan\phi = \frac{gT}{u} = \frac{g\sqrt{2h/g}}{u} = \frac{\sqrt{2gh}}{u}$