Question
What are the angles that give the same range for a projectile launched at 60 degrees, neglecting air resistance?
Answers
GPT-4o mini
To find the angles that give the same range for a projectile launched at \(60^\circ\), we can utilize the properties of projectile motion.
The range \(R\) of a projectile launched with an angle \(\theta\) and an initial speed \(v_0\) is given by the formula:
\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]
where \(g\) is the acceleration due to gravity.
For a launch angle of \(60^\circ\):
\[
R_{60} = \frac{v_0^2 \sin(120^\circ)}{g}
\]
Since \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\), we can express the range at \(60^\circ\) as:
\[
R_{60} = \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g}
\]
To find other angles that give the same range, we set the range equation equal for an unknown angle \(\theta\):
\[
R_\theta = \frac{v_0^2 \sin(2\theta)}{g}
\]
We need to find \(\theta\) such that:
\[
\frac{v_0^2 \sin(2\theta)}{g} = \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g}
\]
Cancelling the \( \frac{v_0^2}{g} \) terms, we have:
\[
\sin(2\theta) = \frac{\sqrt{3}}{2}
\]
The general solutions for \(\sin\) are:
\[
2\theta = 60^\circ + k \cdot 360^\circ \quad \text{or} \quad 2\theta = 120^\circ + k \cdot 360^\circ
\]
Dividing by 2 to solve for \(\theta\):
1. \(\theta = 30^\circ + k \cdot 180^\circ\)
2. \(\theta = 60^\circ + k \cdot 180^\circ\)
For \(k = 0\), we have the specific angles:
- \(30^\circ\)
- \(60^\circ\) (the original angle)
For \(k = 1\):
- \(30^\circ + 180^\circ = 210^\circ\)
- \(60^\circ + 180^\circ = 240^\circ\)
However, angles greater than \(90^\circ\) are generally not considered in the context of projectile motion, so we discard \(210^\circ\) and \(240^\circ\).
In summary, the angles that give the same range as \(60^\circ\) are:
\[
\theta = 30^\circ \quad \text{and} \quad \theta = 60^\circ
\]
The range \(R\) of a projectile launched with an angle \(\theta\) and an initial speed \(v_0\) is given by the formula:
\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]
where \(g\) is the acceleration due to gravity.
For a launch angle of \(60^\circ\):
\[
R_{60} = \frac{v_0^2 \sin(120^\circ)}{g}
\]
Since \(\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\), we can express the range at \(60^\circ\) as:
\[
R_{60} = \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g}
\]
To find other angles that give the same range, we set the range equation equal for an unknown angle \(\theta\):
\[
R_\theta = \frac{v_0^2 \sin(2\theta)}{g}
\]
We need to find \(\theta\) such that:
\[
\frac{v_0^2 \sin(2\theta)}{g} = \frac{v_0^2 \cdot \frac{\sqrt{3}}{2}}{g}
\]
Cancelling the \( \frac{v_0^2}{g} \) terms, we have:
\[
\sin(2\theta) = \frac{\sqrt{3}}{2}
\]
The general solutions for \(\sin\) are:
\[
2\theta = 60^\circ + k \cdot 360^\circ \quad \text{or} \quad 2\theta = 120^\circ + k \cdot 360^\circ
\]
Dividing by 2 to solve for \(\theta\):
1. \(\theta = 30^\circ + k \cdot 180^\circ\)
2. \(\theta = 60^\circ + k \cdot 180^\circ\)
For \(k = 0\), we have the specific angles:
- \(30^\circ\)
- \(60^\circ\) (the original angle)
For \(k = 1\):
- \(30^\circ + 180^\circ = 210^\circ\)
- \(60^\circ + 180^\circ = 240^\circ\)
However, angles greater than \(90^\circ\) are generally not considered in the context of projectile motion, so we discard \(210^\circ\) and \(240^\circ\).
In summary, the angles that give the same range as \(60^\circ\) are:
\[
\theta = 30^\circ \quad \text{and} \quad \theta = 60^\circ
\]