A body projected horizontally from top of tower. When its velocity is at an angle $60^0$ with horizontal  its speed is ${\mathrm{V}}_1$ and  when it is moving at an angle $30^0$ with horizontal its speed is $V_{2 }$then  is ${\mathrm{V}}_1:V_{2 }=$         

As the acceleration along horizontal is zero, horizontal component of velocity is constant

When the velocity makes 60^o with horizontal. ${\mathrm{V}}_{\mathrm{x}}={\mathrm{V}}_1\cos {60}^{\mathrm{o}}\Rightarrow {\mathrm{V}}_1=\frac{{\mathrm{V}}_{\mathrm{x}}}{\cos {60}^{\mathrm{o}}}$

When the velocity makes 30^o with horizontal. ${\mathrm{V}}_{\mathrm{x}}={\mathrm{V}}_2\cos {30}^{\mathrm{o}}\Rightarrow {\mathrm{V}}_2=\frac{{\mathrm{V}}_{\mathrm{x}}}{\cos {30}^{\mathrm{o}}}$

${\mathrm{V}}_1:{\mathrm{V}}_{2 }=\sqrt{3}:1$