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karebareyo
07.10.2019 •
Physics
While sitting on a tree branch 10 m above the ground, you drop a chestnut. when the chestnut has fallen 2.5 m, you throw a second chestnut straight down. what initial speed must you give the second chestnut if they are both to reach the ground at the same time?
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Ответ:
10.5 m/s
Explanation:
For the first chestnut:
y₀ = 10 m
v₀ = 0 m/s
a = -9.8 m/s²
y = y₀ + v₀ t + ½ at²
y = 10 + (0) t + ½ (-9.8) t²
y = 10 − 4.9t²
When y = 7.5:
7.5 = 10 − 4.9t²
t = 5/7
When y = 0:
0 = 10 − 4.9t²
t = 10/7
For the second chestnut:
y₀ = 10 m
y = 0 m
a = -9.8 m/s²
t = 10/7 s − 5/7 s = 5/7 s
y = y₀ + v₀ t + ½ at²
0 = 10 + v₀ (5/7) + ½ (-9.8) (5/7)²
0 = 10 + 5/7 v₀ − 2.5
v₀ = -10.5
The second chestnut must be thrown downwards at 10.5 m/s.
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