The body, thrown vertically upward, fell to the ground after 6 seconds. To what maximum height did the body rise and what is its speed at the moment of impact on the ground? Acceleration of gravity to be taken equal to 10 m / s ^ 2
Task data: t (total duration of movement) = 6 s.
Constants: g (acceleration due to gravity) = 10 m / s2.
1) Since the total duration of movement of a given body in the air is 6 s, it took 3 s to rise and fall: t1 = t2 = t / 2 = 3 s.
2) The initial speed of the throw (speed at the moment of impact): V1 = V0 + a * t1 = 0 = V0 – g * t1, whence we express: V0 = g * t1 = 10 * 3 = 30 m / s or V2 = 0 + g * t2 = 30 m / s.
3) Lift height: hmax = S = V0 * t1 + a * t1 ^ 2/2 = V0 * t1 – g * t1 ^ 2/2 = 30 * 3 – 10 * 3 ^ 2/2 = 45 m.
Answer: The body has risen 45 m; its speed before impact was 30 m / s.
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