Projectile motion is a two-dimensional kinematics model in which an object is launched with an initial speed
\(v_0\)
at an angle
\(\theta\)
above the horizontal from an initial height
\(y_0\).
The standard model assumes no air resistance and constant gravity downward. Under those conditions, the horizontal and vertical motions can be treated separately and then combined into a full trajectory.
Resolve the launch speed into components
The first step is to split the initial velocity into horizontal and vertical components.
Horizontal component.
\[
\begin{aligned}
v_{0x} &= v_0 \cos\theta
\end{aligned}
\]
Vertical component.
\[
\begin{aligned}
v_{0y} &= v_0 \sin\theta
\end{aligned}
\]
These two components have very different roles. Since there is no horizontal acceleration in the basic model, the horizontal speed remains constant. The vertical motion, however, is accelerated downward by gravity.
Parametric equations of motion
The position of the projectile is described by the pair
\((x(t), y(t))\).
The horizontal motion is uniform:
\[
\begin{aligned}
x(t) &= v_{0x}\, t
\end{aligned}
\]
The vertical motion is uniformly accelerated:
\[
\begin{aligned}
y(t) &= y_0 + v_{0y}\, t - \tfrac{1}{2} g t^2
\end{aligned}
\]
The horizontal and vertical velocity components are
\[
\begin{aligned}
v_x(t) &= v_{0x} \\
v_y(t) &= v_{0y} - g t
\end{aligned}
\]
So in the ideal projectile model,
\(v_x\)
is constant, while
\(v_y\)
decreases linearly with time.
Time of flight
The projectile lands when it reaches ground level,
\(y = 0\).
To find the flight time
\(T\),
set
\(y(T)=0\):
\[
\begin{aligned}
0 &= y_0 + v_{0y} T - \tfrac{1}{2} g T^2
\end{aligned}
\]
Solving the quadratic gives the nonnegative flight time:
\[
\begin{aligned}
T &= \frac{v_{0y} + \sqrt{v_{0y}^2 + 2 g y_0}}{g}
\end{aligned}
\]
This reduces to the familiar
\(T = 2v_{0y}/g\)
when the projectile starts and lands at the same height, that is, when
\(y_0 = 0\).
Maximum height
The maximum height occurs at the apex, where the vertical velocity becomes zero:
\[
\begin{aligned}
0 &= v_{0y} - g t_* \\
t_* &= \frac{v_{0y}}{g}
\end{aligned}
\]
Substituting that time into the vertical position equation gives
\[
\begin{aligned}
y_{\max} &= y_0 + \frac{v_{0y}^2}{2g}
\end{aligned}
\]
This formula makes the physical dependence clear: a larger vertical launch component raises the apex, while a larger gravity lowers it.
Horizontal range
The horizontal range is the horizontal position at landing:
\[
\begin{aligned}
R &= x(T) = v_{0x} T
\end{aligned}
\]
Because
\(v_{0x}\)
is constant, the full effect of changing gravity on the range comes through the time of flight.
Worked prompt example
Take
\(v_0 = 20\,\mathrm{m\,s^{-1}}\),
\(\theta = 45^\circ\),
\(y_0 = 1.5\,\mathrm{m}\),
and
\(g = 9.81\,\mathrm{m\,s^{-2}}\).
Step 1. Resolve the speed into components.
\[
\begin{aligned}
v_{0x} &= 20 \cos 45^\circ \approx 14.14\,\mathrm{m\,s^{-1}} \\
v_{0y} &= 20 \sin 45^\circ \approx 14.14\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
Step 2. Solve the flight time.
\[
\begin{aligned}
T &= \frac{14.14 + \sqrt{14.14^2 + 2(9.81)(1.5)}}{9.81} \\
&\approx 2.98\,\mathrm{s}
\end{aligned}
\]
Step 3. Compute the maximum height.
\[
\begin{aligned}
y_{\max} &= 1.5 + \frac{14.14^2}{2(9.81)} \\
&\approx 11.7\,\mathrm{m}
\end{aligned}
\]
Step 4. Compute the range.
\[
\begin{aligned}
R &= 14.14 \times 2.98 \\
&\approx 42.2\,\mathrm{m}
\end{aligned}
\]
Small numerical differences from other rounded examples come from whether intermediate values are rounded early or retained at full precision.
Why the animation shows multiple vectors
The animation is especially useful because it separates the total velocity vector from its two components. The green horizontal arrow shows the constant
\(v_x\),
the red vertical arrow shows the changing
\(v_y\),
and the diagonal total-velocity arrow combines both. This makes the motion easier to interpret:
| Vector |
Behavior |
Meaning |
| \(v_x\) |
Constant |
Uniform horizontal motion |
| \(v_y\) |
Decreases linearly |
Upward climb, then downward fall |
| \(\vec v\) |
Changes in both size and direction |
Total instantaneous velocity |
| \(\vec a\) |
Constant downward |
Gravity only |
What the graphs mean
The time graphs complement the trajectory plot:
| Graph |
Shape |
Reason |
| \(x(t)\) |
Linear |
Horizontal velocity is constant |
| \(y(t)\) |
Parabolic |
Vertical motion has constant acceleration |
| \(v_x(t)\) |
Horizontal line |
No horizontal acceleration |
| \(v_y(t)\) |
Straight line with slope \(-g\) |
Gravity changes vertical velocity uniformly |
Summary
| Quantity |
Formula |
Meaning |
| Horizontal launch component |
\(v_{0x} = v_0 \cos\theta\) |
Initial horizontal speed |
| Vertical launch component |
\(v_{0y} = v_0 \sin\theta\) |
Initial vertical speed |
| Horizontal position |
\(x(t) = v_{0x} t\) |
Uniform horizontal motion |
| Vertical position |
\(y(t) = y_0 + v_{0y} t - \tfrac{1}{2} g t^2\) |
Accelerated vertical motion |
| Flight time |
\(T = \dfrac{v_{0y} + \sqrt{v_{0y}^2 + 2 g y_0}}{g}\) |
Time until the projectile reaches \(y=0\) |
| Maximum height |
\(y_{\max} = y_0 + \dfrac{v_{0y}^2}{2g}\) |
Height at the apex |
| Range |
\(R = v_{0x} T\) |
Horizontal landing distance |
