Free fall and general vertical motion under gravity are both one-dimensional constant-acceleration problems. The only difference is the initial conditions. In pure free fall, the object is often dropped from rest, so \(v_0 = 0\). In the more general vertical-motion case, the object can start upward or downward with any initial velocity \(v_0\).
Sign convention
This calculator uses the standard convention
upward positive.
That means:
| Quantity |
Meaning |
| \(y\) |
Height above the ground level \(y = 0\) |
| \(v_0 > 0\) |
Initially upward |
| \(v_0 < 0\) |
Initially downward |
| \(a = -g\) |
Gravity always points downward |
With this convention, gravity is a constant negative acceleration:
\[
\begin{aligned}
a &= -g
\end{aligned}
\]
Main equations
Since the acceleration is constant, the standard kinematic equations apply:
Height as a function of time.
\[
\begin{aligned}
y(t) &= y_0 + v_0 t - \tfrac{1}{2} g t^2
\end{aligned}
\]
Velocity as a function of time.
\[
\begin{aligned}
v(t) &= v_0 - g t
\end{aligned}
\]
Acceleration.
\[
\begin{aligned}
a(t) &= -g
\end{aligned}
\]
These formulas are enough to solve all three combined calculator modes: values at a chosen time, apex, and impact with the ground.
At time \(t\)
In this mode, the calculator substitutes the chosen time directly into the formulas for \(y(t)\) and \(v(t)\). It also computes the vertical change
\(\Delta y = y(t) - y_0\)
and the distance fallen from the starting height,
\(s(t) = \max(0, y_0 - y(t))\).
The distance fallen is especially useful in the free-fall case. For a drop from rest,
\(v_0 = 0\),
so the formulas simplify to
\[
\begin{aligned}
y(t) &= y_0 - \tfrac{1}{2} g t^2 \\
v(t) &= -g t \\
s(t) &= y_0 - y(t) = \tfrac{1}{2} g t^2
\end{aligned}
\]
Apex or maximum height
The apex occurs when the vertical velocity becomes zero.
\[
\begin{aligned}
0 &= v_0 - g t_{\max}
\end{aligned}
\]
Solving gives
\[
\begin{aligned}
t_{\max} &= \frac{v_0}{g}
\end{aligned}
\]
This only makes sense as a future apex when
\(v_0 > 0\).
If the object is not thrown upward, there is no later apex; the highest point for
\(t \ge 0\)
is simply the initial point.
The maximum height is then
\[
\begin{aligned}
H_{\max} &= y_0 + v_0 t_{\max} - \tfrac{1}{2} g t_{\max}^2
\end{aligned}
\]
Substituting
\(t_{\max} = v_0/g\)
gives the well-known compact form
\[
\begin{aligned}
H_{\max} &= y_0 + \frac{v_0^2}{2g}
\end{aligned}
\]
Impact with the ground
Ground impact occurs when the height becomes zero:
\[
\begin{aligned}
0 &= y_0 + v_0 t - \tfrac{1}{2} g t^2
\end{aligned}
\]
Rearranging into quadratic form:
\[
\begin{aligned}
g t^2 - 2 v_0 t - 2 y_0 &= 0
\end{aligned}
\]
Applying the quadratic formula gives
\[
\begin{aligned}
t &= \frac{v_0 \pm \sqrt{v_0^2 + 2 g y_0}}{g}
\end{aligned}
\]
The physically meaningful impact time is the nonnegative root that corresponds to the future instant when the object reaches
\(y = 0\).
Once the impact time is known, the impact velocity is
\[
\begin{aligned}
v_{\text{hit}} &= v_0 - g t_{\text{hit}}
\end{aligned}
\]
Worked prompt example
Use the sample values
\(y_0 = 20\,\mathrm{m}\),
\(v_0 = 15\,\mathrm{m\,s^{-1}}\),
and
\(g = 9.81\,\mathrm{m\,s^{-2}}\).
Find the impact time.
Step 1. Set the ground height equal to zero.
\[
\begin{aligned}
0 &= 20 + 15 t - \tfrac{1}{2}(9.81)t^2
\end{aligned}
\]
Step 2. Use the impact formula.
\[
\begin{aligned}
t_{\text{hit}} &= \frac{15 + \sqrt{15^2 + 2(9.81)(20)}}{9.81}
\end{aligned}
\]
Step 3. Evaluate numerically.
\[
\begin{aligned}
t_{\text{hit}} &\approx 4.04\,\mathrm{s}
\end{aligned}
\]
Step 4. Compute the impact velocity.
\[
\begin{aligned}
v_{\text{hit}} &= 15 - 9.81(4.04) \\
&\approx -24.6\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The negative sign means the object is moving downward at impact, exactly as expected from the upward-positive convention.
What the graphs and animation show
The position graph
\(y(t)\)
is a parabola because the acceleration is constant. The velocity graph
\(v(t)\)
is a straight line with slope
\(-g\).
The acceleration graph is a horizontal line because gravity is constant. The animation helps connect those graphs to the physical motion: the ball rises or falls, the green arrow shows velocity direction and size, and the red arrow always points downward because gravity always acts downward.
Summary
| Quantity |
Formula |
Meaning |
| Height |
\(y(t) = y_0 + v_0 t - \tfrac{1}{2} g t^2\) |
Vertical position at time \(t\) |
| Velocity |
\(v(t) = v_0 - g t\) |
Vertical velocity at time \(t\) |
| Acceleration |
\(a = -g\) |
Constant downward acceleration |
| Apex time |
\(t_{\max} = v_0/g\) |
Only for \(v_0 > 0\) |
| Maximum height |
\(H_{\max} = y_0 + v_0^2/(2g)\) |
Peak height above the ground |
| Impact time |
\(t = \dfrac{v_0 \pm \sqrt{v_0^2 + 2gy_0}}{g}\) |
Choose the physical nonnegative root |
