Question

a particle is projected with velocity v so that its horizontal range twice the greatest height attained .The horizontal range is​

Answer 1

Answer:

Let H = greatest height attained

R = Range

So by the given condition

R=2H

Let θ be the angle

Again we have

R = 4Hcotθ

⇒cotθ= 1/2

So

sinθ= 2/√5

cosθ= 1/√5

So, the range

= 2v² sinθ cosθ / g

= 2v² ×2/√5× 1/√5/ g

= 4v²/5 g