Projectile Range

What this chapter uses

This chapter uses two results of the previous part. (1.2.1) gives the Velocity under a constant acceleration. (1.1.7) gives the position under a constant acceleration.

From launch to landing

Assumptions

This section uses two assumptions. First, the projectile lands at the height from which it was launched (level ground). Second, air resistance is zero.

The launch velocity is stated in polar form: a Launch speed and a Launch angle above the horizontal. (1.1.3) converts the launch velocity into a horizontal component $\explain{motion.launch-speed}{v_0}\cos(\explain{motion.launch-angle}{\theta})$ and a vertical component $\explain{motion.launch-speed}{v_0}\sin(\explain{motion.launch-angle}{\theta})$. The Gravitational acceleration is vertical. So the horizontal velocity component is constant during the flight. The vertical velocity component decreases, is zero at the highest point, and is negative after the highest point. Diagram 2.1.1 shows the steps of the derivation:

Diagram 2.1.1Steps of the range derivation. From the launch conditions to the range, through the two velocity components and the time of flight.

The time of flight is the time from launch to landing. The vertical position is zero at launch and at landing. Setting the vertical position to zero and solving for the time gives a time of flight of $2\explain{motion.launch-speed}{v_0}\sin(\explain{motion.launch-angle}{\theta})/\explain{gravity}{g}$. The Range is the horizontal distance from the launch point to the landing point. For a constant horizontal velocity component, the range is that component multiplied by the time of flight. Substituting both expressions gives the range:

$$\begin{aligned} \explain{motion.range}{R} &= \explain{motion.launch-speed}{v_0}\cos(\explain{motion.launch-angle}{\theta})\cdot\frac{2\explain{motion.launch-speed}{v_0}\sin(\explain{motion.launch-angle}{\theta})}{\explain{gravity}{g}}\\ &= \dfrac{\explain{motion.launch-speed}{v_0}^{2}\sin(2\explain{motion.launch-angle}{\theta})}{\explain{gravity}{g}} \quad\text{double-angle identity} \end{aligned}$$

(2.1.1)

Solving (2.1.1) for the Launch speed gives the launch speed that produces a given range. Three symbols of the result are under the square root sign, and each of them still opens its explanation:

$$\explain{motion.launch-speed}{v_0} = \sqrt{\explain{motion.range}{R}\,\explain{gravity}{g}\,\csc\!\left(2\explain{motion.launch-angle}{\theta}\right)}$$

(2.1.2)

Equation (2.1.2) uses $\csc(2\explain{motion.launch-angle}{\theta}) = 1/\sin(2\explain{motion.launch-angle}{\theta})$. It is defined only for $\sin(2\explain{motion.launch-angle}{\theta}) > 0$, i.e., for $0^{\circ} < \explain{motion.launch-angle}{\theta} < 90^{\circ}$.

Caution

(2.1.1) is correct only if the launch point and the landing point are at the same height and air resistance is zero [1].

Launching from a height

If the launch point is higher than the landing point, (2.1.1) does not apply. Let Launch height be the height of the launch point above the landing plane. The time of flight is then the positive solution of a quadratic equation for the vertical position [2]. The range is the horizontal velocity component multiplied by this time of flight:

$$\explain{motion.range}{R} = \dfrac{\explain{motion.launch-speed}{v_0}\cos(\explain{motion.launch-angle}{\theta})}{\explain{gravity}{g}}\left[\explain{motion.launch-speed}{v_0}\sin\left(\explain{motion.launch-angle}{\theta}\right) + \sqrt{\left(\explain{motion.launch-speed}{v_0}\sin\left(\explain{motion.launch-angle}{\theta}\right)\right)^{2} + 2\explain{gravity}{g}\explain{launch-height}{h}}\right]$$

(2.1.3)

For Launch height $= 0$, the square root in (2.1.3) equals $\explain{motion.launch-speed}{v_0}\sin(\explain{motion.launch-angle}{\theta})$. So the bracket equals $2\explain{motion.launch-speed}{v_0}\sin(\explain{motion.launch-angle}{\theta})$, and (2.1.3) reduces to (2.1.1).

(2.1.3) also shows how this engine sizes delimiters. Three scalable pairs are nested: the parentheses of the squared group are inside the square root, and the square root is inside the outer bracket. The square root is as tall as a fraction, and the outer bracket must be taller than the square root. Stock KaTeX lets a \left … \right pair be slightly shorter than its content. Then all three pairs get the same glyph, and the reader cannot see the nesting. This engine uses a patched KaTeX that makes every pair strictly taller than its content. So each level is visibly larger than the level inside it, and the source contains no manual size command such as \Bigl or \biggl.

Worked example: try it

The explorer in Figure 2.1.1 evaluates (2.1.1) for a launch speed and a launch angle that you choose. If (2.1.1) is on the screen, a reference to it highlights the equation. If the equation is off the screen, the reference opens a preview of it.

Figure 2.1.1Projectile trajectory explorer. The path from launch to landing; the dot marks the landing point.

Controls of Figure 2.1.1

Move either slider to change the trajectory and the range.

Launch speed 20 m/sExact launch speed in meters per second

Launch angle 30°Exact launch angle in degrees above the horizontal

$$\explain{motion.range}{R} = \dfrac{\explain{motion.launch-speed}{v_0}^2\sin(2\explain{motion.launch-angle}{\theta})}{\explain{gravity}{g}} = \dfrac{ \htmlData{lab-slot=v0}{20.0} ^2\sin(2\times \htmlData{lab-slot=theta}{30} \text{°})}{ \htmlData{lab-slot=g}{9.81} } = \htmlData{lab-slot=range}{35.3} \text{ m}$$

For a fixed launch speed, a larger launch angle gives a higher trajectory, but not always a larger range. The range is the horizontal velocity component multiplied by the time of flight. A larger launch angle increases the time of flight but decreases the horizontal velocity component.

The next chapter plots the range against the launch angle for one launch speed, in Figure 2.2.1. That plot shows which launch angle gives the largest range.

A reference to Figure 2.1.1 scrolls back to the explorer.

Sampled flight data

The explorer shows one trajectory at a time. Table 2.1.1 instead lists the state of one flight at eleven values of Time, from launch to landing. The flight has a Launch speed of $45\ \text{m/s}$ and a Launch angle of $55^\circ$. Each row gives the Position, the Velocity, and the Acceleration from the constant-acceleration equations of the previous part. Each row also gives the Kinetic energy and the Potential energy per unit Mass:

Table 2.1.1Sampled flight data. Position, velocity, acceleration and specific mechanical energy through one flight, from launch to landing.

Time (seconds after launch)$\explain{motion.time}{t}\ \text{(s)}$	Position (meters)$\explain{motion.position}{x}\ \text{(m)}$	Position (meters)$\explain{motion.position}{y}\ \text{(m)}$	Velocity (meters per second)$\explain{motion.velocity}{v_x}\ \text{(m/s)}$	Velocity (meters per second)$\explain{motion.velocity}{v_y}\ \text{(m/s)}$	Acceleration (meters per second squared)$\explain{motion.acceleration}{a_x}\ \text{(m/s}^{2}\text{)}$	Acceleration (meters per second squared)$\explain{motion.acceleration}{a_y}\ \text{(m/s}^{2}\text{)}$	Velocity (meters per second)$\lvert\explain{motion.velocity}{\mathbf{v}}\rvert\ \text{(m/s)}$	Launch angle (degrees above the horizontal)$\explain{motion.launch-angle}{\theta}\ \text{(°)}$	Kinetic energy (joules) / Mass (kilograms)$\explain{kinetic-energy}{\mathrm{KE}}/\explain{mass}{m}\ \text{(J/kg)}$	Potential energy (joules) / Mass (kilograms)$\explain{potential-energy}{\mathrm{PE}}/\explain{mass}{m}\ \text{(J/kg)}$
0.00	0.00	0.00	25.81	36.86	0.00	-9.80	45.00	55.00	1012.50	0.00
0.75	19.36	24.89	25.81	29.51	0.00	-9.80	39.21	48.83	768.58	243.92
1.50	38.72	44.27	25.81	22.16	0.00	-9.80	34.02	40.65	578.68	433.82
2.25	58.07	58.13	25.81	14.81	0.00	-9.80	29.76	29.85	442.80	569.70
3.00	77.43	66.49	25.81	7.46	0.00	-9.80	26.87	16.12	360.94	651.56
3.76	97.05	69.33	25.81	0.01	0.00	-9.80	25.81	0.03	333.10	679.40
4.50	116.15	66.65	25.81	-7.24	0.00	-9.80	26.81	-15.67	359.30	653.20
5.25	135.51	58.47	25.81	-14.59	0.00	-9.80	29.65	-29.47	439.51	572.99
6.00	154.87	44.77	25.81	-21.94	0.00	-9.80	33.87	-40.36	573.74	438.76
6.75	174.22	25.56	25.81	-29.29	0.00	-9.80	39.04	-48.61	762.00	250.50
7.52	194.17	0.00	25.81	-36.86	0.00	-9.80	45.00	-55.00	1012.50	0.00

Eleven columns are enough to make Table 2.1.1 scroll horizontally on most screens. Each row agrees with the results of this chapter:

• The Gravitational acceleration appears only in the $\explain{motion.acceleration}{a_y}$ column. The $\explain{motion.acceleration}{a_x}$ column is zero in every row.
• The $\explain{motion.velocity}{v_x}$ column has the same value in every row.
• The sum of the last two columns is $1012.50\ \text{J/kg}$ in every row. This sum is the mechanical energy per unit Mass. The mechanical energy is constant because air resistance is zero.
• At $\explain{motion.time}{t} = 3.76\ \text{s}$, the $\explain{motion.velocity}{v_y}$ column is approximately zero. So this row is the highest point of the flight.
• The last row is the landing. For this launch speed and launch angle, (2.1.1) gives $\explain{motion.range}{R} \approx 194\ \text{m}$, and the $\explain{motion.position}{x}$ column gives $194.17\ \text{m}$.

Knowledge check 2.1.1 Projectile range

1. DirectQuestion 1Not answered

References

1. Ling et al., University Physics Volume 1 (2016). ch. 4, “Projectile Motion”. https://openstax.org/details/books/university-physics-volume-1 draft ↩
2. Urone & Hinrichs, College Physics 2e (2022). ch. 3, “Projectile Motion”, Example 3.5. https://openstax.org/details/books/college-physics-2e draft ↩