Two stones are thrown from the same point, on the ground, in the same direction. The first has an initial velocity of modulo $20$ $\text{m/s}$ and forms an angle of $60^{\circ}$ with the horizontal, while for the other stone, this angle is $30^{\circ}$. The modulus of the second stone's initial velocity, so they both have the same range, is?
Note: disregard air resistance
Attempt: Since they have the same range, this means that, by symmetry, the throw range $A$ up to the maximum height of the first stone will be equal to the maximum range of the second stone. That is, by the equation of velocity for the vertical velocity of the first pebble: $0 = 10\sqrt{3}-10t \implies t= \sqrt{3}$, with $t$ being the instant in which they reach their maximum height.
So we have that $$v_{o_2} \cdot \frac{\sqrt{3}}{2} \sqrt{3}= 20 \frac{1}{2} \cdot \sqrt{3} \implies v_{0_2}=\frac{ 20\sqrt{3}}{3}$$
But the answer is $20$ $\text{m/s}$. I wonder why I'm wrong
If the particle is projected at speed $V$ and at angle $\theta$ to the horizontal, the range over horizontal ground is $$\frac{V^2\sin2\theta}{g}$$
Since $\sin120=\sin 60$, both $V$ values are the same.
The error you seem to be making is in assuming "the throw range A up to the maximum height of the first stone will be equal to the maximum range of the second stone". Why is this?
Note that the time of flight over horizontal ground is $$\frac{2V\sin\theta}{g}$$ and the time to maximum height is half of this. The times of flight are not directly related, only the ranges are equal.