Question

A particle is projected with a velocity v
such that its range on the horizontal
plane is twice the greatest height
attained by it. The range of the projectile
is (where g is acceleration due to gravity)​

Answer 1

Explanation:

Given, Range, R=2H

But R=4Hcotθ

⇒cotθ=1/2

From triangle we can say that,

sinθ= 52and cocos

So, the range of the projectile

R= g2v 2 sinθcosθ

= g2v 2 × 52 × 51

= 5g4v 2

Hope it will help you