The projectile fired from the gun at an angle of α = 30 degrees to the horizon was twice at the same height

The projectile fired from the gun at an angle of α = 30 degrees to the horizon was twice at the same height h: after a time t1 = 10 s and t2 = 50 s after the shot. Determine the initial speed υ0 and height h.

When the body is launched at an angle to the horizontal, the horizontal component of its velocity does not change, and the vertical component changes according to the equation for uniformly accelerated motion:
Vy = Voy – g * t,
Where
Voy = V0 * sin (30 °) = V0 / 2 – initial vertical component of speed,
g = 10m / s ^ 2 – free fall acceleration;
According to the law of conservation of energy at the same height h, the body will have the same kinetic energy, that is, the speed of the body at two points at the height h will be the same. But at the first point Vy will be directed upward, and at the second point it will be directed downward.
Vy1 = V0y – g * t1,
Vy2 = V0y – g * t2,
Vy1 = – Vy2;
Hence:
V0y – g * t1 = – (V0y – g * t2);
V0y = g * (t1 + t2) / 2 = 10 * 30 = 300m / s;
V0 = 2 * V0y = 600m / s;
From the equation for uniformly accelerated motion:
h = V0y * t1 – g * (t1 ^ 2) / 2 = 300 * 10 – 10 * 50 = 2500m.