Question

A particle is thrown with velocity v such that its range on horizontal plane is twice to the maximum height attained. Its range will be?

• A. $\frac{2v^2}{3g}$
• B. $\frac{4v^2}{3g}$
• C. $\frac{4v^2}{5g}$
• D. $\frac{v^2}{7g}$

Answer 1

The correct option is C $\frac{4v^2}{5g}$

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)