Question

A PROJECTILE IS THROWN WITH A INITIAL VELOCITY OF V=ai^+bj^,IF THE RANGE OF THE PROJECTILE IS DOUBLE ,THE MAXIMUM HEIGHT REACHED BY IT THEN-

Answer 1

v = a i + b j
       b = dy/dt           a = dx/dt
       b/a = dy/dx = tan θ

Range = R  = 2 H
Max Height = H 
H/R = 1/4 * tan θ

So tan θ = 2

tan θ = b/a = 2      =>  b = 2 a

Answer 2

b=. 2a. is write answers..............