Question

A particle is projected with a velocity v so that its range
on a horizontal plane is twice the greatest height attained. Show that the range
is
4v^2/58​

Answer 1

Answer:

Given, Range, R=2H

But R=4Hcotθ

⇒cotθ=

2

1

From triangle we can say that,

sinθ=

5

2

and cosθ=

5

1

So, the range of the projectile

R=

g

2v

2

sinθcosθ

=

g

2v

2

×

5

2

×

5