A small stone thrown from a flat horizontal surface of the earth at an angle to the horizon, in one second

A small stone thrown from a flat horizontal surface of the earth at an angle to the horizon, in one second it was at a height of 15m. Determine the maximum flight range of the stone. The minimum stone speed is 5m / s.

It is known from the problem statement that a small stone thrown from a flat horizontal surface of the earth at an angle to the horizon, after one second (t = 1 s), turned out to be at a height of h = 15 meters. In this case, the stone participates in two types of movement: uniform horizontal flight by inertia and vertical movement under the influence of gravity. At each moment of time, the speed of the stone is found as the vector sum of the horizontal and vertical components of the speed. The horizontal component of the velocity is equal to the minimum stone velocity Vx = 5 m / s. The vertical component of the velocity changes and its initial value is determined from the formula Vо = (h + g ∙ t² / 2): t, where the acceleration of gravity g = 9.8 m / s². We get that Vou = (15 + 9.8 ∙ 1² / 2): 1; Vou = 19.9 m / s. Then you can determine the maximum flight altitude: H = V²ou: (2 ∙ g); H = 19.9²: (2 ∙ 9.8); H ≈ 20.2 m. To determine the ascent time tp to this height, we will use the formula: t²п = 2 ∙ Н / g; tp ≈ 2 s. Hence, it is possible to determine the maximum flight range of a stone L = 2 ∙ Vх ∙ t; L = 2 ∙ 5 ∙ 2; L ≈ 20 m.
Answer: the maximum flight range of a stone is ≈ 20 meters.