CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
60o
loader
C
45o
loader
D
0o
loader

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 B

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image