Question

the max range of a projectile is 2 / root 3 times its actual range . what is the angle of projection for the actual range ?

Answer 1

Actual Horizontal Range (R) of a projectile is given by:
R = (u² sin2@)/g.
where @ is angle of projection.
(u = projection velocity , g = gravity )
Maximum range (Rmax)of a projectile is given by:
Rmax= u²/g. ( at projection angle=45° ).
Now according to question:
u²/g =( 2/√3 )×( u² sin2@)/g.
on solving we get,
sin2@ = √3/2.
=> 2@ = 60°.
=> @ = 60°/2 = 30°.
Hence angle of projection of actual range is 30°.

Answer 2

u2/g = u2 sin 2/g x 2/√3 or, sin 2= √3/2 or

2 = 60° or, θ = 30°

corresponds to theta