for what angle horizontal projection is maximum a)90 degree (b) 45 degree (c) 60 degree (d) 0 degr

Horizontal range , X = v_o² sin 2 θ / g

From above eqn , it is clear that horizonatal range depends on the value of velocity of projection and angle of projection. For fixed v_{o ,} range will be maximum if sin 2 θ is max. i.e.

sin 2 θ = 1 = sin 90^o

or, 2 θ = 90^o

or, θ = 45^o