Time of Flight

What this formula says

A projectile is launched from the ground with speed $u$ at an angle $\theta$ above the horizontal. It follows a curved path and lands back on the same level ground. The total time it spends in the air is:

$T = \frac{2u \sin\theta}{g}$

where $g$ is the acceleration due to gravity ($\approx 9.8 \text{ m/s}^2$, acting downward).

Setting up the problem

Projectile motion is two independent motions happening simultaneously:

• Horizontal: constant velocity — no force acts horizontally (ignoring air resistance)
• Vertical: constant downward acceleration $g$ — gravity acts throughout

We only need the vertical motion to find the time of flight. The horizontal motion tells us nothing about when the projectile lands.

The initial velocity has two components:

• Horizontal: $u_x = u\cos\theta$
• Vertical: $u_y = u\sin\theta$

The projectile is launched from the ground ($s = 0$) and lands back on the ground ($s = 0$). So the net vertical displacement at landing is zero.

Derivation

Apply the second equation of motion to the vertical direction. Taking upward as positive:

• Initial vertical velocity: $u_y = u\sin\theta$
• Acceleration: $-g$ (downward, so negative)
• Displacement at landing: $s = 0$

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

$0 = u\sin\theta \cdot T - \frac{1}{2}gT^2$

Factor out $T$:

$0 = T\left(u\sin\theta - \frac{1}{2}gT\right)$

This gives two solutions:

$T = 0 \quad \text{or} \quad u\sin\theta - \frac{1}{2}gT = 0$

$T = 0$ is the moment of launch — not what we want. The second solution gives:

$\frac{1}{2}gT = u\sin\theta$

$\boxed{T = \frac{2u\sin\theta}{g}}$

Physical interpretation

The time of flight is exactly twice the time to reach maximum height. This makes sense by symmetry — under constant gravity, the upward journey and the downward journey take equal time. The projectile spends the first half of $T$ going up, and the second half coming down.

Time to reach maximum height (where $v_y = 0$):

$0 = u\sin\theta - g \cdot t_{\text{top}}$

$t_{\text{top}} = \frac{u\sin\theta}{g}$

And indeed $T = 2 \times t_{\text{top}}$.

How $\theta$ affects time of flight

$T = \frac{2u\sin\theta}{g}$

$\sin\theta$ increases from $0°$ to $90°$, so:

Assumptions

This derivation assumes:

• Launch and landing are at the same height
• No air resistance
• $g$ is constant (valid near Earth's surface)

If the projectile lands at a different height, the net vertical displacement is not zero, and the formula changes.