Projectiles $$A$$ and $$B$$ are thrown at angles of $$45°$$ and $$60°$$ with vertical respectively from top of a 400 m high tower. If their times of flight are same, the ratio of their speeds of projection $$v_A : v_B$$ is:

Given,

Two Projectiles A and B  are thrown with an angle  of $$45°$$ and $$60°$$ with vertical

In the question it is specified that the angle is with respect to the vertical 

For the Time of flight (T) to be same their vertical velocities must be equal

So

As it is with respect to vertical the vertical velocities are:-

For A $$v_A=v_a\cos\theta\ _A$$ ($$\cos\theta\ \ as\ \ \theta\ \ is\ with\ respect\ to\ vertical$$)

and For B $$v_B=v_b\cos\theta\ _A$$

Now these both should be equal
so ,

$$v_a\cos\left(45^{\circ\ }\right)=v_b\cos\left(60^{\circ\ }\right)$$

$$v_a\times\ \frac{1}{\sqrt{\ 2}}=v_b\times\ \frac{1}{2}$$

$$\frac{v_a}{v_b}=\ \frac{\sqrt{\ 2}}{2}=\frac{1}{\sqrt{\ 2}}$$