Two projectiles A and B thrown with speeds in the ratio $1:\sqrt{2}$acquired the same heights.  If A is thrown at an angle of 45° with the horizontal, the angle of projection of B will be 

For projectile A 
Maximum height, ${\mathrm{H}}_{\mathrm{A}}=\frac{{\mathrm{u}}_{\mathrm{A}}^2{\sin }^2{45}^{\circ }}{2\mathrm{g}}$
For projectile B
Maximum height,  ${\mathrm{H}}_{\mathrm{B}}=\frac{{\mathrm{u}}_{\mathrm{B}}^2{\sin }^2\theta }{2\mathrm{g}}$
As per question,  ${\mathrm{H}}_{\mathrm{A}}={\mathrm{H}}_{\mathrm{B}}$
$\frac{{\mathrm{u}}_{\mathrm{A}}^2{\sin }^2{45}^{\circ }}{2\mathrm{g}}=\frac{{\mathrm{u}}_{\mathrm{B}}^2{\sin }^2\theta }{2\mathrm{g}}$
$\begin{array}{rl} & \Rightarrow \frac{{\sin }^2\theta }{{\sin }^2{45}^{\circ }}=\frac{u_A^2}{u_B^2}\Rightarrow {\sin }^2\theta ={\left(\frac{u_A}{u_B}\right)}^2{\sin }^2{45}^{\circ }\\ & \Rightarrow {\sin }^2\theta ={\left(\frac{1}{\sqrt{2}}\right)}^2{\left(\frac{1}{\sqrt{2}}\right)}^2[\because \frac{u_A}{u_B}=\frac{1}{\sqrt{2}}\text{(Given)}]\end{array}$
$\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{2}\right)={30}^{\circ }$