Three-Dimensional Launch Motion Workshop

Introduction

A launch problem in three dimensions asks how an object moves when its initial position and velocity are known and gravity is the only acceleration. The horizontal and vertical components can behave differently, but they combine into one position vector and one velocity vector.

The reliable approach is to set the coordinate directions first, represent gravity as a vector, and use the same time value in each component. The resulting formulas give position and velocity at any time during the modeled flight. Speed is then the magnitude of the velocity, while acceleration remains the constant gravity vector.

Set the Coordinate System and Gravity

Let \(x\) and \(y\) be horizontal coordinates and let \(z\) point vertically upward. In this coordinate system, gravity points in the negative \(z\)-direction. If \(g\) is the positive magnitude of gravitational acceleration, then the acceleration vector is

\mathbf{a}(t)=\langle 0,0,-g\rangle.

Near Earth’s surface, a common approximation is \(g=9.8\text{ m/s}^2\). Some problems use \(g=10\text{ m/s}^2\) to simplify arithmetic. Use the value specified in the problem, and do not treat \(g\) itself as a negative number: the negative sign in the vector specifies the direction.

Definition: A three-dimensional launch under constant gravity is a motion with specified initial position \(\mathbf{r}_0=\langle x_0,y_0,z_0\rangle\), specified initial velocity \(\mathbf{v}_0=\langle v_{x0},v_{y0},v_{z0}\rangle\), and constant acceleration \(\mathbf{a}=\langle 0,0,-g\rangle\), where \(g>0\). Time \(t=0\) is the launch instant.

The Motion Formulas

For this model, the position and velocity at time \(t\geq 0\) are

\mathbf{r}(t) = \left\langle x_0+v_{x0}t,\; y_0+v_{y0}t,\; z_0+v_{z0}t-\frac12gt^2 \right\rangle,

\mathbf{v}(t) = \left\langle v_{x0},\; v_{y0},\; v_{z0}-gt \right\rangle, \qquad \mathbf{a}(t)=\langle 0,0,-g\rangle.

The speed is the nonnegative magnitude of the velocity:

\|\mathbf{v}(t)\| = \sqrt{v_{x0}^{\,2}+v_{y0}^{\,2}+(v_{z0}-gt)^2}.

These formulas assume that gravity is the only acceleration and that its magnitude and direction remain constant over the time interval considered. They do not include effects such as air resistance. If a problem specifies a different coordinate orientation or a different constant acceleration vector, the corresponding acceleration components must be used instead.

Why the Formulas Work

Position, velocity, and acceleration are related by \(\mathbf{v}(t)=\mathbf{r}'(t)\) and \(\mathbf{a}(t)=\mathbf{v}'(t)\). Since acceleration is constant, integrating it from the launch time to time \(t\), then applying the initial velocity, gives

\mathbf{v}(t) = \mathbf{v}_0+\int_0^t\langle 0,0,-g\rangle\,ds = \mathbf{v}_0+\langle 0,0,-gt\rangle.

Integrating this velocity and applying the initial position gives the position formula. In particular, the vertical component accumulates the integral of \(v_{z0}-gs\), while the two horizontal velocity components remain constant.

\mathbf{r}(t) = \mathbf{r}_0+\int_0^t\mathbf{v}(s)\,ds = \left\langle x_0+v_{x0}t,\; y_0+v_{y0}t,\; z_0+v_{z0}t-\frac12gt^2 \right\rangle.

This also explains why the coordinate calculations can be handled separately: each component of the vector equation follows the corresponding scalar equation. The coordinates still describe one three-dimensional motion, and their components must all be evaluated at the same time.

Worked Launch Problems

Worked Example: A launch with gravity of ten metres per second squared

An object starts at \(\mathbf{r}_0=\langle 1,-2,3\rangle\text{ m}\) with velocity \(\mathbf{v}_0=\langle 4,1,6\rangle\text{ m/s}\). Take \(g=10\text{ m/s}^2\). Find position, velocity, speed, and acceleration at \(t=2\text{ s}\).

The acceleration is \(\langle 0,0,-10\rangle\text{ m/s}^2\). For position, substitute \(t=2\) in each coordinate:

\begin{aligned} x(2)&=1+(4)(2)=9\text{ m},\\ y(2)&=-2+(1)(2)=0\text{ m},\\ z(2)&=3+(6)(2)-\frac12(10)(2^2)\\ &=3+12-20=-5\text{ m}. \end{aligned}

Thus \(\mathbf{r}(2)=\langle 9,0,-5\rangle\text{ m}\). The velocity components are

\mathbf{v}(2) = \langle 4,1,6-(10)(2)\rangle = \langle 4,1,-14\rangle\text{ m/s}.

The speed is

\|\mathbf{v}(2)\| = \sqrt{4^2+1^2+(-14)^2} = \sqrt{16+1+196} = \sqrt{213}\text{ m/s}.

The acceleration at \(2\text{ s}\) is still \(\langle 0,0,-10\rangle\text{ m/s}^2\). To check the results, differentiating the position function gives \(\mathbf{r}'(t)=\langle 4,1,6-10t\rangle\), which yields the stated velocity at \(t=2\); differentiating again gives the stated acceleration. At \(t=0\), the position and velocity are the specified initial vectors.

Worked Example: A launch that reaches its highest point at the requested time

An object starts at \(\mathbf{r}_0=\langle 2,-1,0\rangle\text{ m}\) with velocity \(\mathbf{v}_0=\langle 3,4,19.6\rangle\text{ m/s}\). Use \(g=9.8\text{ m/s}^2\). Find its position, velocity, speed, and acceleration at \(t=2\text{ s}\).

The position is calculated component by component:

\begin{aligned} x(2)&=2+(3)(2)=8\text{ m},\\ y(2)&=-1+(4)(2)=7\text{ m},\\ z(2)&=0+(19.6)(2)-\frac12(9.8)(2^2)\\ &=39.2-19.6=19.6\text{ m}. \end{aligned}

Therefore, \(\mathbf{r}(2)=\langle 8,7,19.6\rangle\text{ m}\). The velocity is

\mathbf{v}(2) = \langle 3,4,19.6-(9.8)(2)\rangle = \langle 3,4,0\rangle\text{ m/s}.

Its speed is \(\sqrt{3^2+4^2+0^2}=\sqrt{9+16}=5\text{ m/s}\). The acceleration is \(\langle 0,0,-9.8\rangle\text{ m/s}^2\). The vertical velocity is zero at this time, so the object is at the top of its vertical motion; its horizontal velocity is not zero, so its total speed is still \(5\text{ m/s}\).

For a check, the functions are \(\mathbf{r}(t)=\langle 2+3t,-1+4t,19.6t-4.9t^2\rangle\) and \(\mathbf{v}(t)=\langle 3,4,19.6-9.8t\rangle\). Differentiating the first gives the second, and differentiating the second gives \(\langle 0,0,-9.8\rangle\). Substituting \(t=0\) recovers both initial vectors.

Worked Example: A horizontal launch from above the ground

An object is launched horizontally from \(\mathbf{r}_0=\langle 1,-2,20\rangle\text{ m}\) with \(\mathbf{v}_0=\langle 3,4,0\rangle\text{ m/s}\). Let \(g=9.8\text{ m/s}^2\). Find position, velocity, speed, and acceleration at \(t=1.5\text{ s}\).

The horizontal coordinates are

x(1.5)=1+(3)(1.5)=5.5\text{ m}, \qquad y(1.5)=-2+(4)(1.5)=4\text{ m}.

The vertical coordinate is

z(1.5) = 20+(0)(1.5)-\frac12(9.8)(1.5^2) = 20-4.9(2.25) = 20-11.025 = 8.975\text{ m}.

Thus \(\mathbf{r}(1.5)=\langle 5.5,4,8.975\rangle\text{ m}\). The velocity at this time is

\mathbf{v}(1.5) = \langle 3,4,0-(9.8)(1.5)\rangle = \langle 3,4,-14.7\rangle\text{ m/s}.

Its speed is

\|\mathbf{v}(1.5)\| = \sqrt{3^2+4^2+(-14.7)^2} = \sqrt{9+16+216.09} = \sqrt{241.09}\text{ m/s}.

The acceleration is \(\langle 0,0,-9.8\rangle\text{ m/s}^2\). For a final check, the position function is \(\langle 1+3t,-2+4t,20-4.9t^2\rangle\); its derivative is \(\langle 3,4,-9.8t\rangle\), which at \(t=1.5\) gives the velocity above. Differentiating once more gives the acceleration, and \(t=0\) gives the supplied initial position and velocity.

Reading the Results Correctly

The position vector describes where the object is relative to the chosen origin. A negative coordinate is not inherently an error; it means the object lies in the negative direction of that coordinate axis. In the first example, \(z=-5\text{ m}\) is the mathematical position predicted by the model at the requested time. Whether the object would actually reach the ground sooner depends on the physical setup, which the given launch data alone do not specify.

Velocity has direction as well as magnitude. A negative vertical component, such as \(-14.7\text{ m/s}\), means the object is moving downward. Speed has no direction and cannot be negative: it is the square root of the sum of the squared velocity components. Acceleration is not the same as velocity; under the stated model, its vertical component remains \(-g\) even when vertical velocity is zero.

Key takeaway: Keep the vector quantities distinct. Position and velocity change with time, speed is the magnitude of velocity, and the acceleration under constant gravity is the fixed vector \(\langle 0,0,-g\rangle\).

A Common Sign Mistake

A frequent error is to use \(+gt\) for the vertical velocity while also choosing \(z\) upward. For example, with \(v_{z0}=0\), \(g=9.8\text{ m/s}^2\), and \(t=1\text{ s}\), the incorrect calculation \(v_z=0+9.8(1)=9.8\text{ m/s}\) predicts upward motion. But gravity points downward in the chosen coordinate system. The correct calculation is \(v_z=0-9.8(1)=-9.8\text{ m/s}\).

The position sign must agree with this velocity rule: the vertical displacement due to gravity is \(-\frac12gt^2\), not \(+\frac12gt^2\). A useful check is to differentiate the proposed vertical position. The derivative must equal the proposed vertical velocity, and its derivative must equal \(-g\).

Check Your Understanding

Use \(z\) positive upward and assume gravity is the only acceleration. For each problem, find the requested quantities and check your functions by differentiation.

An object starts at \(\langle 0,1,2\rangle\text{ m}\) with velocity \(\langle 2,-1,5\rangle\text{ m/s}\). With \(g=10\text{ m/s}^2\), find position and velocity at \(t=1\text{ s}\), then find speed and acceleration.
For an initial vertical velocity of \(14.7\text{ m/s}\) and \(g=9.8\text{ m/s}^2\), what is the vertical velocity after \(1.5\text{ s}\)? Explain what its sign means.
Why can the speed be positive when the vertical velocity is zero?
Write the position and velocity functions for initial position \(\langle 2,0,10\rangle\) and initial velocity \(\langle 0,3,0\rangle\), using \(g=9.8\text{ m/s}^2\).
Which acceleration vector applies throughout a launch in this model, including at the instant when the object is at the top of its path?