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?

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.

c. newton's third law of motion