Question

prove that there are two angles of projection for the same horizontal range ​

Answer 1

To prove:

2 angles of projection for which projectiles have same range.

Proof :

General expression of the range of a projectile is:

$R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

• R is range , u is initial velocity, g is gravity and $\theta$ is angle of projection.

Now, let's assume that we have a complimentary angle as 90° - $\theta$. For that, the range will be :

$R_{2} = \dfrac{ {u}^{2} \sin \{2 ( {90}^{ \circ} - \theta) \} }{g}$

$\implies R_{2} = \dfrac{ {u}^{2} \sin ( {180}^{ \circ} - 2\theta) }{g}$

$\implies R_{2} = \dfrac{ {u}^{2} \sin ( 2\theta) }{g}$

$\implies R_{2} = R$

So, for complimentary angles of projection, the range will be same (when initial velocity remains same).