A body thrown vertically upward, 3 s after the start of movement, had a speed of 10 m / s.

A body thrown vertically upward, 3 s after the start of movement, had a speed of 10 m / s. What is the maximum height of the body relative to the place of throw?

These tasks: t (considered ascent time) = 3 s; V (speed at the moment in time) = 10 m / s.

Constants: by condition g (gravitational acceleration) = 9.8 m / s2.

1) The initial speed of the throw of a given body: V = Vn + a * t = Vn – g * t, whence we express: Vn = V + g * t = 10 + 9.8 * 3 = 39.4 m / s.

2) The maximum height of a given body relative to the place of throw: hmax = (Vk ^ 2 – Vn ^ 2) / 2a = (Vn ^ 2 – 0) / 2g = 39.4 ^ 2 / (2 * 9.8) ≈ 79, 2 m.

Answer: With respect to the place of the throw, the given body was able to rise 79.2 m.