The body, thrown vertically upward, reached a height of 45m with an interval of 8s.

The body, thrown vertically upward, reached a height of 45m with an interval of 8s. Determine the modulus of the initial body velocity.

Data: h1 (height passed by the thrown body twice) = 45 m; Δt (time elapsed between passes of the considered height) = 8 s.

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

1) Duration of ascent after passing the considered height: t1 = t2 = Δt / 2 = 8/2 = 4 s.

2) The difference between the considered height and the maximum: Δh = g * t1 ^ 2/2 = 10 * 42/2 = 80 m.

3) Maximum height: h = h1 + Δh = 45 + 80 = 125 m.

4) Initial speed module: V0 = √ (2 * g * h) = √ (2 * 10 * 125) = 50 m / s.

Answer: The body was thrown at a speed of 50 m / s.