The body thrown vertically upward fell to the ground 4 seconds after the throw, to determine how much the potential

The body thrown vertically upward fell to the ground 4 seconds after the throw, to determine how much the potential energy of the body will increase in the second second of the movement if the body weight is 8.4 kg. air resistance is negligible.

Task data: t (total duration of movement) = 4 s; m (the mass of the thrown body) = 8.4 kg.

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

1) Since the fall of the thrown body occurred in 4 s, its ascent upward lasted: tv = t / 2 = 4/2 = 2 s.

2) Initial speed: V0 = g * tv = 10 * 2 = 20 m / s.

3) The increase in potential energy in the second second: ΔEp = En2 – En1 = m * g * h2 – m * g * h1 = m * g * (h2 – h1) = m * g * (V0 * t ^ 2 – g * t2 ^ 2/2 – V0 * t1 + g * t1 ^ 2/2) = 8.4 * 10 * (20 * 2 – 10 * 2 ^ 2/2 – 20 * 1 + 10 * 1 ^ 2/2) = 420 J.

Answer: In the second second, the potential energy increased by 420 J (in the first second the body rose by 15 m, in the second – by 5 m, the total ascent height was 20 m).