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For angles of projection of a projectile at angles $(45^\circ - \theta)$ and $(45^\circ + \theta)$, the horizontal ranges described by the projectile are in the ratio of :
A
1 : 1
B
2 : 3
C
1 : 2
D
2 : 1
Detailed Solution
Horizontal range: $R = \dfrac{u^2\sin 2\alpha}{g}$
$R_1 = \dfrac{u^2\sin 2(45^\circ - \theta)}{g} = \dfrac{u^2\sin(90^\circ - 2\theta)}{g} = \dfrac{u^2\cos 2\theta}{g}$
$R_2 = \dfrac{u^2\sin 2(45^\circ + \theta)}{g} = \dfrac{u^2\sin(90^\circ + 2\theta)}{g} = \dfrac{u^2\cos 2\theta}{g}$
$\dfrac{R_1}{R_2} = \dfrac{\cos 2\theta}{\cos 2\theta} = 1$
In other words, $(45^\circ - \theta)$ and $(45^\circ + \theta)$ are complementary angles (their sum is $90^\circ$), and for complementary angles of projection the range is the same.
So the ranges are in the ratio 1 : 1.
$R_1 = \dfrac{u^2\sin 2(45^\circ - \theta)}{g} = \dfrac{u^2\sin(90^\circ - 2\theta)}{g} = \dfrac{u^2\cos 2\theta}{g}$
$R_2 = \dfrac{u^2\sin 2(45^\circ + \theta)}{g} = \dfrac{u^2\sin(90^\circ + 2\theta)}{g} = \dfrac{u^2\cos 2\theta}{g}$
$\dfrac{R_1}{R_2} = \dfrac{\cos 2\theta}{\cos 2\theta} = 1$
In other words, $(45^\circ - \theta)$ and $(45^\circ + \theta)$ are complementary angles (their sum is $90^\circ$), and for complementary angles of projection the range is the same.
So the ranges are in the ratio 1 : 1.