A stone thrown from the surface of the earth at an angle α = 30 ° to the horizon, twice visited the same

A stone thrown from the surface of the earth at an angle α = 30 ° to the horizon, twice visited the same height h after a time t₁ = 3 s and t₂ = 5 s after the start of the movement. Find the initial speed of the stone υ₀. Acceleration of free fall is taken equal to g = 10m / s². Neglect air resistance.

Let’s use the formula S = V0t + at ^ 2/2 where a is the acceleration, v0 is the initial speed, t is the time.

Then the y coordinate will change according to the law: y (t) = vy0t-gt ^ 2/2.
Since the height is the same, we get:
y (3) = vyo * 3-10 * 9/2 = y (5) = vyo * 5-10 * 25/2
2vy0 = 17 * 5
vy0 = 42.5 m / s
Since vy0 = v0 * sinα we get
v0 = vy0 / sinα = 42.5 / (1/2) = 2 * 42.5 = 83 m / s