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yanaberezhnaya
@yanaberezhnaya
August 2021
1
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С какой высоты упало тело, если за последнюю секунду свободного падения оно полетело 1/3 всего пути?
Ответ 148м
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Путь за t с:
S(t) = g*t² / 2 = 5*t²
Путь за (t-1) с:
S(t-1) = g*(t-1)² / 2 = 5*(t-1)²
Тогда:
ΔS = S(t) - S(t-1) = 5*(t² - (t-1)²) = 5*(2*t-1)
S/3 = 5*(2*t-1)
5*t² / 3 = 5*(2*t-1)
t² -6*t+3=0
t ≈ 5,45 с
H = g*t²/2 = 5*(5,45)² ≈ 148 м
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Answers & Comments
Verified answer
Путь за t с:S(t) = g*t² / 2 = 5*t²
Путь за (t-1) с:
S(t-1) = g*(t-1)² / 2 = 5*(t-1)²
Тогда:
ΔS = S(t) - S(t-1) = 5*(t² - (t-1)²) = 5*(2*t-1)
S/3 = 5*(2*t-1)
5*t² / 3 = 5*(2*t-1)
t² -6*t+3=0
t ≈ 5,45 с
H = g*t²/2 = 5*(5,45)² ≈ 148 м