## If, for a given velocity, the maximum range is at a projection angle of 45, then there must be equal ranges for angles above and below this.

Question

If, for a given velocity, the maximum range is at a projection angle of 45, then there must be equal ranges for angles above and below this. Show this explicitly.

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1 year 2021-07-17T03:54:15+00:00 1 Answers 7 views 0

1. Explanation:

The range R of a projectile is given by

$$R = \frac{v_0^2}{g} \sin 2\theta$$

The maximum range $$R_{max}$$ occurs when $$\sin 2\theta = 1\:\text{or}\:\theta = 45°$$. Let $$\alpha$$ be the angle above or below 45°. Now let’s look at the ranges brought about by these angle differences.

Case 1: Angle above 45°

We can write the range as

$$R_+ = \dfrac{v_0^2}{g} \sin 2(45° + \alpha)= \dfrac{v_0^2}{g} \sin (90° + 2\alpha)$$

$$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha + \cos 90° \sin 2\alpha)$$

$$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(1)$$

Case 2: Angle below 45°

We can write the range as

$$R_- = \dfrac{v_0^2}{g} \sin 2(45° – \alpha)= \dfrac{v_0^2}{g} \sin (90° – 2\alpha)$$

$$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha – \cos 90° \sin 2\alpha)$$

$$\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(2)$$

Note that the equations (1) and (2) are identical. Therefore, the ranges are equal if they differ from 45° by the same amount.