During a projectile motion if the maximum height equals the horizontal range,then the angle of projection with the horizontal is :

During a projectile motion if the maximum height equals the horizontal

1.	During a projectile motion if the maximum height equals the horizontal range,then the angle of projection with the horizontal is :
Answer» `TAN^(-1) (1)` `tan^(-1) (2)` `tan^(-1) (3)` `tan^(-1) (4)` Solution :`(u^(2) SIN ^(2) theta)/(2G)= (u^(2) sin 2 theta)/(g)` `:. (sin^(2)theta)/(2) = 2 sin theta COS theta` `:. (sin theta)/( cos theta) = 4` `tan theta = 4` `:. theta = tan^(-1)(4)`