KAHIBARO
Table of Contents
Horizontal reach of a projectile
Projectile range is the horizontal distance a projectile travels from its launch point to its landing point. It is usually denoted by $R$. This chapter focuses only on that horizontal reach, not on the full motion in each direction.
If a projectile is launched from one point and lands back at the same vertical level, the range tells us how far away it lands along the ground.
How range is found
The horizontal range comes from combining two ideas. First, horizontal motion is uniform if air resistance is ignored. Second, the total time in the air determines how long that horizontal motion continues.
So the basic idea is
$$
R = v_{0x} \, T
$$
where $v_{0x}$ is the horizontal component of the initial velocity, and $T$ is the total time of flight.
For an oblique launch with initial speed $v_0$ at angle $\theta$ above the horizontal,
$$
v_{0x} = v_0 \cos\theta
$$
If the projectile lands at the same height from which it was launched, the time of flight is
$$
T = \frac{2v_0\sin\theta}{g}
$$
Substituting into the range formula gives
$$
R = \left(v_0\cos\theta\right)\left(\frac{2v_0\sin\theta}{g}\right)
$$
Using the trigonometric identity $2\sin\theta\cos\theta = \sin 2\theta$,
$$
R = \frac{v_0^2 \sin 2\theta}{g}
$$
For a projectile launched and landing at the same vertical level, with no air resistance,
$$
R = \frac{v_0^2 \sin 2\theta}{g}
$$
This formula is one of the most important results for projectile range.
What the formula tells us
The formula shows that range depends on three things, the launch speed $v_0$, the launch angle $\theta$, and gravitational acceleration $g$.
A larger launch speed gives a larger range because $R$ is proportional to $v_0^2$.
A larger value of $g$ gives a smaller range because stronger gravity brings the projectile down sooner.
The angle dependence comes through $\sin 2\theta$, which leads to an important result about the best launch angle.
Maximum range
Since the largest possible value of $\sin 2\theta$ is 1, the range is greatest when
$$
\sin 2\theta = 1
$$
This happens when
$$
2\theta = 90^\circ
$$
so
$$
\theta = 45^\circ
$$
Therefore, for launch and landing at the same height, the projectile travels the greatest horizontal distance when launched at $45^\circ$.
The maximum possible range for a given launch speed is then
$$
R_{\max} = \frac{v_0^2}{g}
$$
For fixed launch speed and equal launch and landing heights, the maximum range occurs at
$$
\theta = 45^\circ
$$
and the maximum value is
$$
R_{\max} = \frac{v_0^2}{g}
$$
Complementary angles
An interesting result of the range formula is that two different launch angles can produce the same range. This happens because angles that add to $90^\circ$ have the same value of $\sin 2\theta$.
For example, $30^\circ$ and $60^\circ$ give the same range because
$$
\sin(2 \times 30^\circ) = \sin 60^\circ
$$
and
$$
\sin(2 \times 60^\circ) = \sin 120^\circ = \sin 60^\circ
$$
So one low-angle path and one high-angle path can land at the same horizontal distance.
Special cases
Some launch angles give simple results.
| Launch angle | Result for range |
|---|---|
| $\theta = 0^\circ$ | $R = 0$ |
| $\theta = 45^\circ$ | $R$ is maximum |
| $\theta = 90^\circ$ | $R = 0$ |
If the projectile is launched horizontally, it has no upward component, so in the equal-height formula the range becomes zero. That is because this formula assumes launch and landing at the same level and uses a nonzero time from upward and downward motion. Horizontal launch from a height is a different case and belongs with other projectile situations.
Limits of the standard formula
The formula
$$
R = \frac{v_0^2 \sin 2\theta}{g}
$$
works only under specific conditions. The projectile must move without air resistance, and it must land at the same vertical height from which it was launched.
If the landing height is different, the range must be found by using horizontal velocity together with the actual time of flight from the vertical motion.
Do not use
$$
R = \frac{v_0^2 \sin 2\theta}{g}
$$
unless the projectile lands at the same height from which it was launched and air resistance is neglected.
Example
Suppose a ball is launched with speed $v_0 = 20\ \text{m/s}$ at an angle of $30^\circ$. Take $g = 9.8\ \text{m/s}^2$.
Use the range formula:
$$
R = \frac{v_0^2 \sin 2\theta}{g}
$$
Substitute the values:
$$
R = \frac{(20)^2 \sin 60^\circ}{9.8}
$$
Since $\sin 60^\circ \approx 0.866$,
$$
R \approx \frac{400 \times 0.866}{9.8}
$$
$$
R \approx 35.3\ \text{m}
$$
So the projectile lands about $35.3\ \text{m}$ away.
Final idea
Projectile range is the horizontal distance traveled before landing. In the simplest and most common case, equal launch and landing heights with no air resistance, the range is determined by
$$
R = \frac{v_0^2 \sin 2\theta}{g}
$$
This formula reveals two key facts, the greatest range occurs at $45^\circ$, and complementary angles give the same range.