Angle of Velocity at Any Time

What this formula gives

At any instant, the velocity vector points in some direction. The angle $\phi$ that this velocity vector makes with the horizontal is given by:

$\tan\phi = \frac{v_y}{v_x} = \frac{u\sin\theta - gt}{u\cos\theta}$

Derivation

The velocity vector has components:

• Horizontal: $v_x = u\cos\theta$ (constant)
• Vertical: $v_y = u\sin\theta - gt$ (changing)

The angle $\phi$ with the horizontal satisfies:

$\tan\phi = \frac{\text{vertical component}}{\text{horizontal component}} = \frac{v_y}{v_x}$

$\boxed{\tan\phi = \frac{u\sin\theta - gt}{u\cos\theta}}$

This follows directly from the definition of tangent in a right triangle formed by the two components.

How the angle changes during flight

At launch ($t = 0$):

$\tan\phi = \frac{u\sin\theta}{u\cos\theta} = \tan\theta \implies \phi = \theta$

As expected — at launch, the velocity points at angle $\theta$ above horizontal.

As $t$ increases, $v_y = u\sin\theta - gt$ decreases. The angle $\phi$ decreases.

At the top (when $v_y = 0$):

$\tan\phi = 0 \implies \phi = 0°$

The velocity is purely horizontal at the top.

After the top, $v_y$ becomes negative (downward). $\phi$ becomes negative — the velocity now points below the horizontal.

At landing, by symmetry, $\phi = -\theta$ — the velocity makes angle $\theta$ below the horizontal, equal in magnitude to the launch angle.

Finding the time when velocity makes a given angle

If a problem asks: "at what time does the velocity make angle $\alpha$ with the horizontal?", set $\phi = \alpha$ and solve for $t$:

$\tan\alpha = \frac{u\sin\theta - gt}{u\cos\theta}$

$u\cos\theta \cdot \tan\alpha = u\sin\theta - gt$

$gt = u\sin\theta - u\cos\theta\tan\alpha$

$t = \frac{u(\sin\theta - \cos\theta\tan\alpha)}{g}$