Evans Lang
07/24/2024 · High School
is the downward acceleration due to gravity. Find the angle \( \alpha \) for which the range \( R \) is the largest possible. \[ R=\frac{v_{0}^{2}}{g} \sin 2 \alpha \]
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Step-by-step Solution
To find the angle \( \alpha \) that maximizes the range \( R \) given by the equation
\[
R = \frac{v_0^2}{g} \sin 2\alpha,
\]
we need to maximize the term \( \sin 2\alpha \). The sine function reaches its maximum value of 1 when its argument is \( \frac{\pi}{2} \) (or \( 90^\circ \)). Therefore, we set:
\[
2\alpha = \frac{\pi}{2}.
\]
Solving for \( \alpha \):
\[
\alpha = \frac{\pi}{4} = 45^\circ.
\]
Thus, the angle \( \alpha \) that maximizes the range \( R \) is
\[
\boxed{45^\circ}.
\]
Quick Answer
The angle \( \alpha \) that maximizes the range \( R \) is \( 45^\circ \).
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