Newton's Logical Mechanics

Let the balls be kept at height $h$ from the ground.

Now we can find the velocity of ball A by using equations of motion. In this case acceleration of ball A is $g$

So ${ V }_{ A }^{ 2 }=2gh$

Now acceleration of ball B is $g(sin\theta -\mu cos\theta )$ (we can get this by making free body diagrams and then using Newton's 2nd law)

So ${ V }_{ B }^{ 2 }=2g(sin\theta -\mu cos\theta )\frac{h}{sin\theta}$

But

${ V }_{ A }=2\sqrt { 2 } { V }_{ B }$

So $\sqrt { 2gh } =2\sqrt { 2 } \sqrt { 2g(sin\theta -\mu cos\theta )\frac{h}{sin\theta} }$

On squaring both sides we get

$1=8(1-\mu cot\theta)$

or $cot\theta =\frac { 4 }{ 3 }$

So $\theta =arccot { \left( \frac { 4 }{ 3 } \right) } \approx 37$ :)