Question

A particle is projected such that the horizontal range and vertical height are equal. Then the angle of projection is :

Options

• π/6

• tan^-1(4) 

• cot^-1(4) 

• tan^-1(1/4) 

Answer 1

tan^-1(4) 

Explanation - 

Range= height

`("v"_0^2  "sin"  2theta)/"g" = ("v"_0^2  "sin"  2theta)/(2"g")`

`-> 2  "sin" theta = (sin^2theta)/2`

⇒ 4 = tanθ

⇒ θ = tan^-14