Position of Projectile at Any Time

What these equations give

At any time $t$ during flight, the projectile's coordinates are:

$x = u\cos\theta \cdot t$

$y = u\sin\theta \cdot t - \frac{1}{2}gt^2$

Together, $(x, y)$ gives the exact position of the projectile at time $t$. These are parametric equations of the trajectory — $t$ is the parameter.

Derivation

Taking the launch point as the origin, with $x$ horizontal and $y$ vertically upward.

Horizontal position:

Horizontal motion has constant velocity $u\cos\theta$ and zero acceleration:

$x = u_x \cdot t + \frac{1}{2}(0)t^2 = u\cos\theta \cdot t$

$\boxed{x = u\cos\theta \cdot t}$

Vertical position:

Vertical motion has initial velocity $u\sin\theta$ and constant downward acceleration $g$. Taking upward as positive, acceleration is $-g$:

$y = u_y \cdot t + \frac{1}{2}(-g)t^2$

$\boxed{y = u\sin\theta \cdot t - \frac{1}{2}gt^2}$

Both follow directly from the second equation of motion $s = ut + \frac{1}{2}at^2$ applied independently to each direction.

Finding position at a specific time

Example: $u = 20$ m/s, $\theta = 30°$, find position at $t = 1$ s.

$x = 20\cos30° \times 1 = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3 \text{ m}$

$y = 20\sin30° \times 1 - \frac{1}{2}(10)(1)^2 = 20 \times 0.5 - 5 = 10 - 5 = 5 \text{ m}$

Position at $t = 1$ s: $(10\sqrt{3},\ 5)$ m.

From parametric to Cartesian — the trajectory

From the $x$ equation: $t = \frac{x}{u\cos\theta}$

Substitute into the $y$ equation:

$y = u\sin\theta \cdot \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$

$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$

This is the equation of trajectory — a parabola in $x$ and $y$ with time eliminated. See the Equation of Trajectory entry for the full derivation.

Symmetry of the trajectory

The trajectory is symmetric about the highest point (the vertex of the parabola). This means:

• The horizontal distance to the top is exactly half the range: $\frac{R}{2}$
• The time to reach the top is exactly half the total time of flight: $\frac{T}{2}$
• The position at time $t$ and at time $T - t$ are mirror images about the vertical axis through the top