Question

Two projectiles are fired at different angles with the same magnitude of velocity, such that they have the same range. At what angles they might have been projected?

• A. $25^{\circ}$ and $65^{\circ}$
• B. $35^{\circ}$ and $75^{\circ}$
• C. $10^{\circ}$ and $50^{\circ}$
• D. None of these

Answer 1

Answer: (A)

$25^{\circ}$ and $65^{\circ}$

Answer 2

Range of a projectile $(R)$ is given by
$R =\frac{u^{2} \sin 2 \theta}{g}$
where $(u)$ is velocity of projection and $8$ the angle of projection.
Substituting $\left(90^{\circ}-\theta\right)$ in place of $8$ , we get
$R =\frac{u^{2} \sin 2\left(90^{\circ}-\theta\right)}{g}$
$=\frac{u^{2} \sin \left(180^{\circ}-2 \theta\right)}{g}=\frac{u^{2} \sin 2 \theta}{g}$
Hence, horizontal range is same whether the body is projected at $\theta$
or $90^{ \circ}-\theta$.
$\therefore, 25^{\circ}$ or $\left(90^{\circ}-25^{\circ}\right)=65^{\circ}$