KAHIBARO
Table of Contents
Meaning of Time of Flight
In projectile motion, the time of flight is the total time a projectile stays in the air from launch until it lands. It is one of the most useful quantities because it helps determine how long the object moves before returning to the ground or reaching a target height.
Time of flight depends on the vertical motion, not directly on the horizontal motion. This is because gravity acts vertically, so the upward and downward parts of the motion control how long the projectile remains in the air.
Time of Flight for a Projectile Returning to the Same Height
Consider a projectile launched with initial speed $v_0$ at an angle $\theta$ above the horizontal. Its initial vertical velocity is
$$
v_{0y} = v_0 \sin\theta
$$
If the projectile lands at the same vertical level from which it was launched, the vertical position is described by
$$
y = v_{0y} t - \frac{1}{2}gt^2
$$
At landing, the vertical displacement is zero, so
$$
0 = v_{0y} t - \frac{1}{2}gt^2
$$
Factoring out $t$ gives
$$
t\left(v_{0y} - \frac{1}{2}gt\right)=0
$$
One solution is $t=0$, which is the launch moment. The other solution gives the total time of flight:
$$
T = \frac{2v_{0y}}{g}
$$
Since $v_{0y} = v_0\sin\theta$, we can also write
$$
T = \frac{2v_0\sin\theta}{g}
$$
For a projectile launched and landing at the same height,
$$
T = \frac{2v_0\sin\theta}{g}
$$
This formula is valid only when the landing height equals the launch height.
Why the Formula Makes Sense
The projectile rises while gravity slows its upward motion. At the highest point, the vertical velocity becomes zero. Then it falls back down, speeding up under gravity. If the launch and landing heights are the same, the upward time equals the downward time.
So if the time to reach the top is
$$
t_{\text{up}} = \frac{v_{0y}}{g}
$$
then the total time is
$$
T = 2t_{\text{up}} = \frac{2v_{0y}}{g}
$$
This symmetry is the reason for the factor of 2.
Time of Flight from the Vertical Velocity Equation
We can also derive the same result from the vertical velocity equation. At the top of the path,
$$
v_y = v_{0y} - gt
$$
Since $v_y=0$ at the highest point,
$$
0 = v_{0y} - gt
$$
so
$$
t_{\text{up}} = \frac{v_{0y}}{g}
$$
Doubling this gives the total flight time for equal launch and landing heights.
When Launch and Landing Heights Are Different
Sometimes a projectile is launched from a cliff, a platform, or some other elevated point. Then the simple formula above does not work, because the motion is no longer symmetric.
In that case, use the vertical position equation:
$$
y = y_0 + v_{0y}t - \frac{1}{2}gt^2
$$
Set $y$ equal to the landing height and solve the quadratic equation for $t$. The physically meaningful answer is the positive one.
If launch height and landing height are different, do not use
$$
T = \frac{2v_0\sin\theta}{g}
$$
Instead, solve the vertical position equation for $t$.
Dependence on Launch Angle
For a fixed launch speed $v_0$, the time of flight changes with the angle because the vertical component changes.
| Launch angle $\theta$ | Vertical component $v_0\sin\theta$ | Time of flight |
|---|---|---|
| Small | Small | Short |
| Medium | Moderate | Medium |
| Large | Large | Long |
A larger launch angle gives a larger initial vertical velocity, so the projectile stays in the air longer, as long as it lands at the same height.
Special Cases
If the projectile is launched horizontally, then $\theta = 0$ and
$$
v_{0y}=0
$$
The equal height formula gives $T=0$, which only reflects that a projectile launched horizontally does not return to the same height unless it is caught immediately. In practical cases with horizontal launch from a height, the time of flight must be found from vertical fall alone.
If the projectile is launched straight upward, then $\theta = 90^\circ$ and
$$
T = \frac{2v_0}{g}
$$
provided it returns to the same launch height.
Visualizing the Motion
A Simple Example
Suppose a ball is launched with speed $20\,\text{m/s}$ at an angle of $30^\circ$, and it lands at the same height.
First find the vertical component:
$$
v_{0y} = 20\sin 30^\circ = 20 \cdot 0.5 = 10\,\text{m/s}
$$
Then use the time of flight formula:
$$
T = \frac{2v_{0y}}{g} = \frac{2(10)}{9.8} \approx 2.04\,\text{s}
$$
So the ball stays in the air for about $2.04$ seconds.
Common Mistakes
A very common mistake is to use the full launch speed $v_0$ instead of the vertical component $v_0\sin\theta$. Only the vertical component determines the flight time.
Another common mistake is to use the equal height formula when the projectile lands at a different height. In that situation, the correct method is solving the vertical motion equation.
Time of flight comes from vertical motion.
Use $v_{0y} = v_0\sin\theta$, not the horizontal component and not the full speed unless the motion is purely vertical.
Final Idea
Time of flight tells us how long gravity and the initial vertical velocity keep a projectile in the air. For motion that begins and ends at the same height, the result is especially simple:
$$
T = \frac{2v_0\sin\theta}{g}
$$
This formula is central in projectile problems and connects directly to other quantities such as maximum height and range.