Question

# The speed of a projectile when it is at its greatest height is $$\sqrt { { 2 }/{ 5 } }$$ times its speed at half the maximum height. What is its angle of projection?

A
30o
B
60o
C
45o
D
0o

Solution

## The correct option is B $$60^o$$Time taken to reach max height $$=\cfrac{u\sin\theta}{g}$$ Max  height $$=\cfrac{u^2\sin^2 \theta}{2g}$$Half the max height $$=\cfrac{u^2\sin^2\theta}{4g}$$ Horizontal 'v' here $$u\cos\theta$$Vertical velocity at half max height$$=\sqrt{u^2\sin^2\theta-2g(\cfrac{u\sin\theta}{2g})}\\=\cfrac{u\sin\theta}{\sqrt2}$$Velocity at half the max-height$$=\sqrt{u^2\cos^2\theta+\cfrac{u^2\sin^2\theta}{2}}\Rightarrow u\cos\theta=\sqrt{\cfrac{2}{5}}\sqrt{u^2\cos^2\theta+\cfrac{u^2\sin^2\theta}{2}}\\ \Rightarrow u^2\cos^2\theta=\cfrac{2}{5}u^2(1-\sin^2\theta+\cfrac{\sin^2\theta}{2})\\ \Rightarrow 5(1-\sin^2\theta)=2-\sin^2\theta\\ \Rightarrow 3=4\sin^2\theta\Rightarrow \sin\theta=\cfrac{\sqrt3}{2}=\theta=\cfrac{\pi}{3}\Rightarrow (60^\circ)$$Option BMathematics

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