Question

A stone projected with a velocity of u at angle 30° with horizontal reaches maximum height x. When it is projected with velocity u at angle 30° with vertical it reaches maximum height y. Then range of particle is ?

Answer 1

projection at angle Ф = 30° with horizontal : 
    x = u² Sin²Ф /2g = u²/8g
   Range = R = u² Sin 2Ф /g
   x/R = TanФ /4 = 1/4√3
   R = 4√3 x

projection at angle Ф= 30° with vertical (=>  angle π/2 -Ф with horizontal)
     y = u² Cos²Ф/2g = 3u²/8g
     y = 3 x
     Range = u² Sin[2(π/2 - Ф)] / g = u² Sin 2Ф / g
                 = R = 4√3 x = 4 y/√3