Discussion
A projectile is fired with initial velocity \(v_{0}\) making an angle \(\alpha\) with the horizontal axis. The range (R) of the projectile is proportional to, (Neglect the air resistance) :
Answer: D
The range is proportional to \(v_{0}^{2}\) and sin 2\(\alpha\).
The time of flight is \(T = 2v_{0}\sin\alpha/g\), and the horizontal velocity is \(v_{0}\cos\alpha\), so:
\[ \begin{aligned} R &= v_{0}\cos\alpha \cdot \frac{2v_{0}\sin\alpha}{g} \\ &= \frac{v_{0}^{2}\sin 2\alpha}{g} \end{aligned} \]
The range therefore grows as the square of the launch speed and as \(\sin 2\alpha\), which is why it is greatest at 45°. Options with \(v_{0}\) to the first power, or with \(\sin\alpha\), each miss one of the two factors.
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