Modeling Projectile Motion
A projectile’s motion can be described by its horizontal and vertical position as functions of time. In the standard model, horizontal acceleration is zero and vertical acceleration is constant and downward. Once the position functions are known, differentiation gives velocity and acceleration, while the magnitude of velocity gives speed.
The model assumes that air resistance is negligible, the gravitational field is uniform, and the chosen coordinate axes are horizontal and vertical. Take the positive vertical direction to be upward, and let \(g>0\) denote the magnitude of gravitational acceleration. If position is measured in meters and time in seconds, then \(g\) has units of meters per second squared; a common idealized value is \(9.8\text{ m/s}^2\), though convenient examples may use \(10\text{ m/s}^2\).
The constants \(v_{0x}\) and \(v_{0y}\) are signed components: a negative vertical component means the projectile initially moves downward. The model applies over the time interval being studied, provided the assumptions above remain appropriate.
Deriving Velocity and Acceleration
Velocity is the derivative of the position vector, taken component by component. Differentiating the two coordinate functions gives
Acceleration is the derivative of velocity, so
Thus the horizontal velocity stays constant, while the vertical velocity decreases at a constant rate. This is the mathematical meaning of constant downward gravitational acceleration in this model.
The speed can change even though acceleration is constant. In particular, when the projectile rises, its vertical velocity moves toward zero; after the highest point, its vertical velocity is negative and its magnitude increases. The horizontal component remains present in the speed throughout.
Worked Example: Find Position, Velocity, Speed, and Acceleration
Suppose a projectile has position, in meters, given by \(\mathbf{r}(t)=\langle 1+4t,\,2+6t-5t^2\rangle\), where \(t\) is in seconds. Find its position, velocity, speed, and acceleration at \(t=\frac12\) second.
First evaluate the position coordinates:
Therefore the position is \(\langle 3,\frac{15}{4}\rangle\) meters. Differentiate the position functions:
The speed is the magnitude of this velocity:
Differentiating once more gives \(\mathbf{a}(t)=\langle 0,-10\rangle\text{ m/s}^2\). The results check directly: differentiating \(1+4t\) gives \(4\), differentiating \(2+6t-5t^2\) gives \(6-10t\), and differentiating that vertical velocity gives \(-10\). The horizontal acceleration is zero, as required by the model.
Distance and Displacement
For a time interval \([t_1,t_2]\), displacement is the change in position, \(\mathbf{r}(t_2)-\mathbf{r}(t_1)\). Distance traveled is the integral of speed:
Distance is the length of the path actually traveled, so it is not generally equal to the magnitude of displacement. The projectile may travel along a curved path even when its initial and final heights are the same.
When \(v_{0x}\ne 0\), put \(a=|v_{0x}|>0\) and \(u=v_{0y}-gt\). An antiderivative needed for the distance integral is
The logarithm’s argument is positive because \(\sqrt{u^2+a^2}>|u|\). To justify the derivative formula, differentiate the first term inside the brackets: its derivative is \(\sqrt{u^2+a^2}+u^2/\sqrt{u^2+a^2}\). The derivative of the logarithm is \(1/\sqrt{u^2+a^2}\). Combining the terms, including the factor \(a^2\), gives \(2\sqrt{u^2+a^2}\); the outer factor of \(\frac12\) leaves \(\sqrt{u^2+a^2}\).
Since \(du/dt=-g\), substitution into the distance integral gives the exact formula
This expression requires \(g>0\) and \(v_{0x}\ne0\). If \(v_{0x}=0\), speed is \(|v_{0y}-gt|\); its integral can be evaluated directly, splitting the interval where \(v_{0y}-gt\) changes sign if that time lies in the interval.
Worked Distance Calculation
Worked Example: Distance During a Flight That Returns to Its Starting Height
Consider \(\mathbf{r}(t)=\langle 3t,\,10t-5t^2\rangle\), with position in meters and time in seconds, for \(0\le t\le2\). Here \(v_{0x}=3\), \(v_{0y}=10\), and \(g=10\). The projectile starts at \(\langle0,0\rangle\), and at \(t=2\) its position is \(\langle6,0\rangle\). Its displacement is therefore \(\langle6,0\rangle\) meters, with magnitude \(6\) meters.
Its velocity and speed are
Use \(a=3\) and \(u=10-10t\). At the endpoints, \(u(0)=10\) and \(u(2)=-10\), so the distance is
Because \(G_3\) is odd, \(G_3(-10)=-G_3(10)\). This follows from the displayed formula: the first term changes sign, and the logarithm changes sign because \((\sqrt{u^2+9}-u)(\sqrt{u^2+9}+u)=9\). Thus
To check the calculation, differentiating \(G_3(10-10t)\) gives \(-10\sqrt{9+(10-10t)^2}\). The distance integral is therefore its decrease from \(t=0\) to \(t=2\), divided by \(10\), exactly as used above. Also, the distance is greater than the \(6\)-meter displacement magnitude: the speed is at least \(3\text{ m/s}\) throughout the two-second interval and is greater than \(3\text{ m/s}\) except at one instant.
Using the Model to Find the Highest Point and Landing Time
The highest point occurs when the vertical velocity is zero, provided that time lies in the interval of motion. Setting \(v_{0y}-gt=0\) gives \(t=v_{0y}/g\). This identifies a maximum when the projectile is moving upward initially, since vertical velocity is positive before that time and negative afterward. A return to a specified height is found by solving the corresponding equation for \(y(t)\), then retaining only solutions in the time interval of interest.
Worked Example: Maximum Height and Return to Ground
Let \(\mathbf{r}(t)=\langle5t,\,20+15t-5t^2\rangle\), in meters, with \(t\) in seconds and \(t\ge0\). The vertical velocity and acceleration are
At the highest point \(v_y=0\), so \(15-10t=0\), which gives \(t=\frac32\) seconds. Substitution into the vertical position gives
The horizontal position then is \(x(\frac32)=5(\frac32)=\frac{15}{2}\) meters. Thus the highest point is \(\langle\frac{15}{2},\frac{125}{4}\rangle\) meters. The vertical velocity changes from positive to negative at \(t=\frac32\), confirming that this is the maximum height.
To find when the projectile returns to ground level, solve \(20+15t-5t^2=0\). Dividing by \(5\) and factoring gives \((t-4)(t+1)=0\). The solutions are \(t=4\) and \(t=-1\); only \(t=4\) is in the stated interval \(t\ge0\). At that time \(x(4)=20\) meters and \(v_y(4)=15-40=-25\text{ m/s}\). The impact speed is
Checks: \(y(4)=20+60-80=0\), so the position is at ground level, and differentiating the position functions gives the velocity components used above.
A Common Sign Error
With upward chosen as positive, gravity contributes \(-g\) to vertical acceleration. A frequent mistake is to use \(y(t)=y_0+v_{0y}t+\frac12gt^2\) while still describing gravity as downward. For example, with \(y_0=20\), \(v_{0y}=15\), and \(g=10\), that incorrect sign gives \(y(4)=20+60+80=160\) meters and \(v_y(4)=15+40=55\text{ m/s}\). It predicts that the projectile is accelerating upward.
The correction is to use \(y(t)=y_0+v_{0y}t-\frac12gt^2\) and \(v_y(t)=v_{0y}-gt\). Then \(y(4)=20+60-80=0\) and \(v_y(4)=15-40=-25\text{ m/s}\), consistent with a downward return to ground level. A consistent choice of axes and signs is essential.
Why the Model Is Useful
Writing motion as position functions makes several quantities accessible through a single procedure: differentiate to obtain velocity, differentiate again to obtain acceleration, and take the magnitude of velocity to obtain speed. Integrating speed gives distance, while subtracting endpoint positions gives displacement. Keeping these quantities distinct prevents a common conceptual error: equal starting and ending heights do not mean zero distance traveled.
The formulas also provide a check on physical interpretation. The horizontal coordinate changes linearly in time, while the vertical coordinate is quadratic. The vertical velocity changes linearly, and its sign indicates whether the projectile is rising or falling. These conclusions depend on the assumptions of the model, especially constant downward acceleration and negligible air resistance.
Check Your Understanding
Use the constant-acceleration projectile model to answer these questions.
- For \(\mathbf{r}(t)=\langle2+5t,\,3+12t-5t^2\rangle\), find the velocity, acceleration, and speed at \(t=1\) second, including units.
- For a projectile with \(v_{0y}=18\text{ m/s}\) and \(g=9\text{ m/s}^2\), when does its vertical velocity become zero? What does that time represent?
- Explain why the distance traveled over an interval can be greater than the magnitude of displacement.
- State the assumptions behind the position model \(y(t)=y_0+v_{0y}t-\frac12gt^2\).
- If the horizontal velocity is zero, what feature of the speed formula may require splitting the distance integral?
