1. Interpret the given angles carefully

The projectiles are thrown at angles with the vertical:

• For projectile $A$: $45^\circ$ with vertical
• For projectile $B$: $60^\circ$ with vertical

So their components are:

• Horizontal component $= v\sin\theta$ (when angle $\theta$ is with vertical)
• Vertical component $= v\cos\theta$

Thus, $u_{xA}=v_A\sin45^\circ=\frac{v_A}{\sqrt2}, \qquad u_{yA}=v_A\cos45^\circ=\frac{v_A}{\sqrt2}$

$u_{xB}=v_B\sin60^\circ=\frac{\sqrt3}{2}v_B, \qquad u_{yB}=v_B\cos60^\circ=\frac{v_B}{2}$

2. Use the condition: times of flight are same

Both are projected from the same height $h=400\,\text{m}$ and have the same time of flight.

For vertical motion, $0 = h + u_y t - \frac12 gt^2$

Since $h$ and $g$ are same for both, and the time of flight $t$ is same, it follows that their initial vertical components must be same: $u_{yA}=u_{yB}$

So, $\frac{v_A}{\sqrt2}=\frac{v_B}{2}$

Therefore, $2v_A=\sqrt2\,v_B$ $\frac{v_A}{v_B}=\frac{\sqrt2}{2}=\frac{1}{\sqrt2}$

Hence, $v_A:v_B=1:\sqrt2$

3. Check with the range condition

Range $R = u_x \cdot t$.

Given ranges are same and times of flight are same, horizontal components must also be same: $u_{xA}=u_{xB}$

That gives $\frac{v_A}{\sqrt2}=\frac{\sqrt3}{2}v_B$

This would imply $\frac{v_A}{v_B}=\frac{\sqrt6}{2}$

This is inconsistent with the time-of-flight condition.

So the statement "their ranges and times of flight are same" cannot hold simultaneously for the given angles unless there is likely a wording/printing issue in the question. However, among the options, the standard result obtained from equal time of flight for projectiles from the same height is: $v_A:v_B=1:\sqrt2$

This matches one of the options.

4. Evaluate options

• A: $1:2$ ❌
• B: $\sqrt2:1$ ❌
• C: $1:\sqrt2$ ✅
• D: $1:\sqrt3$ ❌

5. Final answer

$\boxed{1:\sqrt2}$

So the correct option is C.