The body is thrown vertically upward. To what height does the body rise if the flight lasts

The body is thrown vertically upward. To what height does the body rise if the flight lasts for 8 seconds? What path does the body cover in 2s, 3s, 4s, 5s?

t = 8 s. V = 0 m / s. g = 9.8 m / s ^ 2. t2 = 2 s. t3 = 3 s. t5 = 5 s. h4 -? h2 -? h3 -? h5 -? A body thrown vertically upward moves under the action of gravity with the acceleration of gravity g. Since the entire flight lasts 8 seconds, the body rises for 4 seconds and the body descends for 4 seconds. t4 = t / 2. According to the definition, acceleration is determined by the formula: g = (V – Vо) / t, where V is the final speed of movement, Vо is the initial speed of movement, t is the time of movement. Since V = 0 m / s the body has stopped, then g = – Vо / t4. The sign “-” means that the acceleration g is directed against the speed of movement Vо, the body is decelerated. Vo = g * t4. Vo = 9.8 m / s ^ 2 * 4 s = 39.2 m / s. Let’s use the formula for uniformly accelerated motion: h4 = Vo * t1 – g * t1 ^ 2/2. h 4 = 39.2 m / s * 4 s – 9.8 m / s ^ 2 * (4 s) ^ 2/2 = 78.4 m. h2 = Vo * t2 – g * t2 ^ 2/2. h2 = 39.2 m / s * 2 s – 9.8 m / s ^ 2 * (2 s) ^ 2/2 = 58.8 m. h3 = Vo * t3 – g * t3 ^ 2/2. h3 = 39.2 m / s * 3 s – 9.8 m / s ^ 2 * (3 s) ^ 2/2 = 73.5 m. h3 = h5. Answer: h2 = 58.8 m, h3 = h5 = 73.5 m, h 4 = 78.4 m