Motion Under Uniform Gravity

Modeling Motion Under Gravity

A ball launched into the air does not move with constant velocity: its velocity changes continuously because gravity accelerates it downward. A useful mathematical model describes that change with one constant vector, then uses the launch position and velocity to determine the motion.

In this tutorial, use a fixed three-dimensional coordinate system with the \(z\)-axis pointing vertically upward. The \(x\)- and \(y\)-axes are horizontal. Near Earth’s surface, and when air resistance is neglected, gravitational acceleration is modeled as constant and downward. This model is suitable for ordinary launch problems over a limited region and time interval; it does not include effects such as wind, drag, or changes in gravity with altitude.

Definition: Motion under uniform gravity is motion with constant acceleration \(\mathbf{a}=-g\mathbf{k}\), where \(g>0\) is the gravitational acceleration magnitude and \(\mathbf{k}\) is the unit vector pointing vertically upward. In the chosen coordinates, \(\mathbf{k}=\langle 0,0,1\rangle\), so \(\mathbf{a}=\langle 0,0,-g\rangle\).

The minus sign is essential: \(g\) is a positive magnitude, but the acceleration points in the negative \(z\)-direction. A commonly used approximate value near Earth’s surface is \(g=9.8\text{ m/s}^2\). The value should be used consistently throughout a calculation.

Position and Velocity from Launch Data

Let \(\mathbf{r}_0\) be the position at launch, and let \(\mathbf{v}_0\) be the velocity at launch. These are the initial conditions at time \(t=0\). Since acceleration is the derivative of velocity, constant acceleration gives \(\mathbf{v}(t)=\mathbf{v}_0+\mathbf{a}t\). Integrating velocity and using the initial position gives the position.

\mathbf{v}(t)=\mathbf{v}_0+\mathbf{a}t, \qquad \mathbf{r}(t)=\mathbf{r}_0+\mathbf{v}_0t+\frac{1}{2}\mathbf{a}t^2.

Substituting the gravitational acceleration vector produces the working formulas for uniform gravity:

\mathbf{v}(t)=\mathbf{v}_0-g t\mathbf{k}, \qquad \mathbf{r}(t)=\mathbf{r}_0+\mathbf{v}_0t-\frac{1}{2}g t^2\mathbf{k}.

Write the launch data as \(\mathbf{r}_0=\langle x_0,y_0,z_0\rangle\) and \(\mathbf{v}_0=\langle v_{x0},v_{y0},v_{z0}\rangle\). The same vector formulas, written component by component, are

\mathbf{r}(t) = \left\langle x_0+v_{x0}t,\; y_0+v_{y0}t,\; z_0+v_{z0}t-\frac{1}{2}gt^2 \right\rangle, $$ $$ \mathbf{v}(t) = \left\langle v_{x0},\; v_{y0},\; v_{z0}-gt \right\rangle.

These expressions show exactly where gravity appears: it contributes a quadratic term to vertical position and a linear term to vertical velocity. The horizontal terms have no gravity contribution in this model. The components are useful for calculating a particular motion; the model itself remains the vector equations above.

Why the Formulas Hold

The formulas follow directly from the definitions of acceleration and velocity, together with the initial conditions. The justification also explains why the factors \(t\) and \(\frac12t^2\) appear.

Start with constant acceleration.
For all times in the model, \(\mathbf{r}''(t)=\mathbf{a}\), where \(\mathbf{a}\) is constant.

Integrate to obtain velocity.
Since \(\mathbf{r}''(t)=\mathbf{a}\), integration gives \(\mathbf{r}'(t)=\mathbf{a}t+\mathbf{C}\), where \(\mathbf{C}\) is a constant vector. The initial velocity condition \(\mathbf{r}'(0)=\mathbf{v}_0\) gives \(\mathbf{C}=\mathbf{v}_0\).

Integrate to obtain position.
Integrating \(\mathbf{r}'(t)=\mathbf{v}_0+\mathbf{a}t\) gives \(\mathbf{r}(t)=\mathbf{v}_0t+\frac12\mathbf{a}t^2+\mathbf{D}\). The condition \(\mathbf{r}(0)=\mathbf{r}_0\) gives \(\mathbf{D}=\mathbf{r}_0\).

Substitute gravity.
Putting \(\mathbf{a}=-g\mathbf{k}\) into the position and velocity formulas gives the stated uniform-gravity model.

As a direct check, differentiating the position formula gives \(\mathbf{r}'(t)=\mathbf{v}_0+\mathbf{a}t\), and differentiating again gives \(\mathbf{r}''(t)=\mathbf{a}\). At \(t=0\), the formulas give \(\mathbf{r}(0)=\mathbf{r}_0\) and \(\mathbf{v}(0)=\mathbf{v}_0\). Thus they satisfy both the acceleration rule and the launch data.

Worked Examples

Worked Example: Finding Position and Velocity at a Given Time

A particle starts at \(\mathbf{r}_0=\langle 1,-2,5\rangle\text{ m}\) with launch velocity \(\mathbf{v}_0=\langle 3,4,6\rangle\text{ m/s}\). Use \(g=9.8\text{ m/s}^2\). Find its position and velocity at \(t=2\text{ s}\).

The acceleration is \(\mathbf{a}=\langle 0,0,-9.8\rangle\text{ m/s}^2\). The position formula gives

\begin{aligned} \mathbf{r}(2) &=\langle 1,-2,5\rangle +\langle 3,4,6\rangle(2) +\frac12\langle 0,0,-9.8\rangle(2^2)\\ &=\langle 1,-2,5\rangle+\langle 6,8,12\rangle+\langle 0,0,-19.6\rangle\\ &=\langle 7,6,-2.6\rangle\text{ m}. \end{aligned}

For velocity,

\begin{aligned} \mathbf{v}(2) &=\langle 3,4,6\rangle+\langle 0,0,-9.8\rangle(2)\\ &=\langle 3,4,6\rangle+\langle 0,0,-19.6\rangle\\ &=\langle 3,4,-13.6\rangle\text{ m/s}. \end{aligned}

Check by differentiating the position function: \(\mathbf{r}'(t)=\langle 3,4,6-9.8t\rangle\text{ m/s}\), which gives \(\mathbf{v}(2)=\langle 3,4,-13.6\rangle\text{ m/s}\). At \(t=0\), position and velocity are the stated launch data. The negative vertical position at \(t=2\) means this mathematical model places the particle below the chosen origin by then; whether it has already struck a surface depends on the physical setting.

Worked Example: A Horizontal Launch from a Height

A ball is launched horizontally from a height of \(19.6\text{ m}\) above level ground with velocity \(\langle 6,0,0\rangle\text{ m/s}\). Take the launch point’s horizontal coordinates to be zero and use \(g=9.8\text{ m/s}^2\). Find the time at which it reaches the ground, its horizontal displacement by then, and its velocity at impact.

The initial position is \(\mathbf{r}_0=\langle 0,0,19.6\rangle\text{ m}\), and the initial vertical velocity is zero. The position formula becomes

\mathbf{r}(t)=\langle 6t,0,19.6-4.9t^2\rangle\text{ m}.

Ground level is \(z=0\). Set the vertical component equal to zero and solve:

19.6-4.9t^2=0,\qquad 4.9t^2=19.6,\qquad t^2=4.

Time is nonnegative, so the impact time is \(t=2\text{ s}\). Substitution gives the impact position \(\mathbf{r}(2)=\langle 12,0,0\rangle\text{ m}\). The horizontal displacement is therefore \(\langle 12,0,0\rangle\text{ m}\), with magnitude \(12\text{ m}\).

The velocity is \(\mathbf{v}(t)=\langle 6,0,-9.8t\rangle\text{ m/s}\), so \(\mathbf{v}(2)=\langle 6,0,-19.6\rangle\text{ m/s}\). Check: the vertical position at \(2\text{ s}\) is \(19.6-4.9(2^2)=19.6-19.6=0\), as required for impact, and differentiating the position gives the velocity just calculated.

Worked Example: An Angled Launch and Its Highest Point

A ball is launched from the origin with velocity \(\mathbf{v}_0=\langle 12,0,16\rangle\text{ m/s}\). Use \(g=9.8\text{ m/s}^2\). Find the time and height at the highest point, then find when it returns to its launch height and its position at that time.

The position and velocity are

\mathbf{r}(t)=\langle 12t,0,16t-4.9t^2\rangle\text{ m}, \qquad \mathbf{v}(t)=\langle 12,0,16-9.8t\rangle\text{ m/s}.

At the highest point, vertical velocity is zero. Solve \(16-9.8t=0\):

t=\frac{16}{9.8}=\frac{80}{49}\text{ s}.

Substitute this time into the vertical position. Since \(4.9=\frac{49}{10}\),

\begin{aligned} z\left(\frac{80}{49}\right) &=16\left(\frac{80}{49}\right)-\frac{49}{10}\left(\frac{80}{49}\right)^2\\ &=\frac{1280}{49}-\frac{49}{10}\cdot\frac{6400}{2401}\\ &=\frac{1280}{49}-\frac{640}{49} =\frac{640}{49}\text{ m} \approx 13.06\text{ m}. \end{aligned}

The ball returns to its launch height when \(z(t)=0\). Factor the vertical position: \(16t-4.9t^2=t(16-4.9t)\). One solution is the launch time \(t=0\); the later solution is \(t=16/4.9=160/49\text{ s}\). At that time,

\mathbf{r}\left(\frac{160}{49}\right) = \left\langle 12\left(\frac{160}{49}\right),0,0 \right\rangle = \left\langle\frac{1920}{49},0,0\right\rangle\text{ m} \approx\langle39.18,0,0\rangle\text{ m}.

Check the return height: \(16(160/49)-4.9(160/49)^2=0\), by the factorization above. The vertical velocity then is \(16-9.8(160/49)=16-32=-16\text{ m/s}\), so the ball is moving downward as it passes the launch height. The formulas also return \(\mathbf{r}(0)=\langle0,0,0\rangle\) and \(\mathbf{v}(0)=\langle12,0,16\rangle\text{ m/s}\).

A Sign Error to Avoid

A frequent mistake is to use \(+g\) in the vertical acceleration while also taking upward as the positive \(z\)-direction. For example, writing \(z(t)=z_0+v_{z0}t+\frac12gt^2\) predicts that a ball released from rest at height \(z_0\) has \(z(t)=z_0+4.9t^2\). That calculation says the ball moves upward, contradicting the chosen direction of gravity.

The correction is to give acceleration its direction: \(a_z=-g\). Therefore the vertical position is \(z(t)=z_0+v_{z0}t-\frac12gt^2\), and the vertical velocity is \(v_z(t)=v_{z0}-gt\). A reliable check is to ask whether a freely falling object with zero initial vertical velocity has negative vertical velocity for \(t>0\). The correct formula gives \(v_z(t)=-gt<0\), as expected.

Using the Model Carefully

The formulas describe the motion only while the assumptions are reasonable. In particular, they treat \(g\) as constant and omit air resistance. A calculated position after an object has hit the ground is a continuation of the mathematical trajectory, not a description of its subsequent physical motion.

When solving a launch problem, identify the coordinate directions and initial data before substituting. Keep units consistent: if positions are in metres and time is in seconds, velocity must be in metres per second and \(g\) in metres per second squared. Finally, check the initial position, the initial velocity, and whether the signs of the vertical terms match the chosen upward direction.

Key takeaway: Under uniform gravity with upward vertical direction, \(\mathbf{a}=\langle0,0,-g\rangle\). The launch data determine the trajectory through \(\mathbf{v}(t)=\mathbf{v}_0+\mathbf{a}t\) and \(\mathbf{r}(t)=\mathbf{r}_0+\mathbf{v}_0t+\frac12\mathbf{a}t^2\).

Check Your Understanding

Use the upward-positive convention and show the equations that support each answer.

A particle has initial position \(\langle 2,1,3\rangle\text{ m}\) and initial velocity \(\langle 4,-2,5\rangle\text{ m/s}\). Write its position and velocity functions using \(g=9.8\text{ m/s}^2\).
Why is gravitational acceleration written as \(\langle0,0,-g\rangle\) when \(g\) itself is positive?
A ball is released from rest at height \(z_0\). What are its vertical position and vertical velocity as functions of time?
A horizontal launch has initial vertical position \(14.7\text{ m}\) and zero vertical velocity. How long does the model predict before the ball reaches \(z=0\), using \(g=9.8\text{ m/s}^2\)?
What derivative and initial-data checks can be used to verify a proposed position function?

What You'll Learn