A projectile's horizontal range over level ground is $\frac{v_0^2\sin 2\theta }{g}$. At what launch angle or angles will the projectile land at half of its maximum possible range?

Verified step by step guidance

1. Start by recalling the formula for the horizontal range of a projectile: $R\; =\; \backslash (\backslash frac\{v\_0^2\; \backslash sin(2\backslash theta)\}\{g\}\backslash )$, where $v\_0$ is the initial velocity, $\backslash (\backslash theta\backslash )$ is the launch angle, and $g$ is the acceleration due to gravity.
2. The maximum possible range occurs when $\backslash (\backslash sin\backslash )(2\backslash (\backslash theta\backslash ))\; =\; 1$, which happens at $2\backslash (\backslash theta\backslash )\; =\; 90^\backslash (\backslash circ\backslash )$ or $\backslash (\backslash theta\backslash )\; =\; 45^\backslash (\backslash circ\backslash )$. Substituting this into the range formula gives the maximum range as $R\_\{\backslash (\backslash text\{max\}\backslash )\}\; =\; \backslash (\backslash frac\{v\_0^2\}\{g\}\backslash )$.
3. To find the angle(s) at which the projectile lands at half the maximum range, set the range equal to half of $R\_\{\backslash (\backslash text\{max\}\backslash )\}$: $\backslash (\backslash frac\{v\_0^2\; \backslash sin(2\backslash theta)\}\{g\}\backslash )\; =\; \backslash (\backslash frac\{1\}\{2\}\backslash )\; \backslash (\backslash cdot\backslash )\; \backslash (\backslash frac\{v\_0^2\}\{g\}\backslash )$.
4. Simplify the equation by canceling $\backslash (\backslash frac\{v\_0^2\}\{g\}\backslash )$ on both sides: $\backslash (\backslash sin\backslash )(2\backslash (\backslash theta\backslash ))\; =\; \backslash (\backslash frac\{1\}\{2\}\backslash )$.
5. Solve for $2\backslash (\backslash theta\backslash )$ using the inverse sine function: $2\backslash (\backslash theta\backslash )\; =\; \backslash (\backslash arcsin\backslash )(\backslash (\backslash frac\{1\}\{2\}\backslash ))$. This gives two possible solutions for $2\backslash (\backslash theta\backslash )$ within one full rotation: $2\backslash (\backslash theta\backslash )\; =\; 30^\backslash (\backslash circ\backslash )$ and $2\backslash (\backslash theta\backslash )\; =\; 150^\backslash (\backslash circ\backslash )$. Divide each by 2 to find the corresponding launch angles: $\backslash (\backslash theta\backslash )\; =\; 15^\backslash (\backslash circ\backslash )$ and $\backslash (\backslash theta\backslash )\; =\; 75^\backslash (\backslash circ\backslash )$.

Key Concepts

Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and is subject to the force of gravity. It can be analyzed in two dimensions: horizontal and vertical. The horizontal motion is uniform, while the vertical motion is influenced by gravitational acceleration. Understanding the principles of projectile motion is essential for determining the range and trajectory of the projectile.

Range of a Projectile

The range of a projectile is the horizontal distance it travels before landing, which is influenced by its initial velocity, launch angle, and the acceleration due to gravity. The formula for the maximum range is given by R = v₀² sin 2θ/g, where v₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. This relationship shows how the angle of launch affects the distance traveled.

Launch Angle

The launch angle is the angle at which a projectile is launched relative to the horizontal. It plays a critical role in determining the range and height of the projectile's trajectory. For maximum range, the optimal launch angle is 45 degrees. To find the angles that yield half of the maximum range, one must analyze the sine function in the range formula, which can yield multiple angles for a given range.