• # question_answer The speed of projectile at its maximum height is half of its initial speed. The angle of projection is A) ${{60}^{o}}$                     B) ${{15}^{o}}$         C) ${{30}^{o}}$            D) ${{45}^{o}}$

Conserving the horizontal component component of the projectile, in context to figure, we get $v=u\cos \theta$ $\frac{u}{2}=u\cos \theta \Rightarrow \cos \theta =\frac{1}{2}$or $\theta ={{60}^{o}}$ Hence, the correction option is (a).