The body is thrown vertically upward. If in the first second of flight the body has risen to a height of h = 15 m

The body is thrown vertically upward. If in the first second of flight the body has risen to a height of h = 15 m, then in the first three seconds of flight it will cover a distance equal to …

Task data: t1 (first flight time) = 1 s; h1 (body height after time t1) = 15 m; t3 (considered flight time) = 3 s.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Initial speed of the throw: h1 = V0 * t1 – g * t1 ^ 2/2, whence we express: V0 = (h1 + g * t1 ^ 2/2) / t1 = (15 + 10 * 1 ^ 2/2) / 1 = 20 m / s.

2) Duration of ascent: tv = V0 / g = 2 s.

3) The time it took to fall: tн = t – tв = 1 s.

4) Distance traveled: S = S1 + S2 = V0 * tv – g * tv ^ 2/2 + g * tn ^ 2/2 = 20 * 2 – 10 * 2 ^ 2/2 + 10 * 12/2 = 40 – 20 + 5 = 25 m.

Answer: In the first three seconds of flight (2 s ascending and 1 s falling), the body covered a distance of 25 m.