Question

a particle is projected with a velocity v such that its range on horizontal plane is twice the maximum height. The range of the projectile is ?

Answer 1

H = (R/4) Tanθ

Given that R = 2H

H = (H/2) Tanθ
Tanθ = 2

If Tanθ = 2 then
Sinθ = 2 / √5
Cosθ = 1 / √5

R = 2u² sinθ cosθ / g
= (2u² × 2 / 5) / g
= 4u² / (5 g)

Range of projectile is 4u² / (5g)

Answer 2

Answer:

Then put the value of sin square theta and u will get the answer as 4u square/5g