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.

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

Answers ( )

    0
    2021-07-17T03:55:33+00:00

    Explanation:

    The range R of a projectile is given by

    [tex]R = \frac{v_0^2}{g} \sin 2\theta[/tex]

    The maximum range [tex]R_{max}[/tex] occurs when [tex]\sin 2\theta = 1\:\text{or}\:\theta = 45°[/tex]. Let [tex]\alpha[/tex] 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

    [tex]R_+ = \dfrac{v_0^2}{g} \sin 2(45° + \alpha)= \dfrac{v_0^2}{g} \sin (90° + 2\alpha)[/tex]

    [tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha + \cos 90° \sin 2\alpha)[/tex]

    [tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(1)[/tex]

    Case 2: Angle below 45°

    We can write the range as

    [tex]R_- = \dfrac{v_0^2}{g} \sin 2(45° – \alpha)= \dfrac{v_0^2}{g} \sin (90° – 2\alpha)[/tex]

    [tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} (\sin 90° \cos 2\alpha – \cos 90° \sin 2\alpha)[/tex]

    [tex]\:\:\:\:\:\:\:= \dfrac{v_0^2}{g} \cos 2\alpha\:\:\:\:\:(2)[/tex]

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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )