Question

For a projectile projected from ground at an angle theta with horizontal gT²=2R/root3 where T is time of flight, R is horizontal range of projectile, g is acceleration due to gravity. The ange of projection theta is

Answer 1

Answer:

30 degrees

Explanation:

Answer 2

Given info : For a projectile projected from ground at an angle theta with horizontal , $gT^2=\frac{2R}{\sqrt{3}}$ . where T is time of flight, R is horizontal range of projectile, g is acceleration due to gravity.

To find : the angle of projection is ..

solution : we know,

• time of flight , T = $\frac{2usin\theta}{g}$
• horizontal range , R = $\frac{u^2sin2\theta}{g}$

now, $gT^2=\frac{2R}{\sqrt{3}}$

⇒ $g\left(\frac{2usin\theta}{g}\right)^2=\frac{2u^2sin2\theta}{g\sqrt{3}}$

⇒ 4√3sin²θ = 2sin2θ

⇒ 4√3sin²θ = 2(2sinθcosθ)

⇒ √3 sinθ/cosθ = 1

⇒ tanθ = 1/√3 = tan30°

⇒ θ = 30°

therefore the angle of projection is 30°.