Question

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?

Answer 1

as the first, after being dropped from a height of 10 m, is approximately 12.2 m/s.

Step-by-step explanation:

the first, we need to use the equations of motion for falling bodies, assuming acceleration due to gravity g =

Let's find the time it takes for the first chestnut to fall the remaining 7.5 m after it has already fallen 2.5 m:

t = √(2d/g), where:

• d is the distance the chestnut falls,
• g is the acceleration due to gravity,
• t is the time.

Substituting the known values gives us:

t = √(2*7.5 m / 9.81 m/s2) = √(15/9.81) s ≈ 1.24 s

The second chestnut must cover the whole 10 m distance. Its motion can be described by:

v = u + gt

To ensure they reach the ground simultaneously, the second chestnut's total fall time must match the first's. We need to find the initial upward velocity u required to achieve this:

u = v - gt = 0 - 9.81 m/s2 * 1.24 s ≈ -12.2 m/s

Answer 2

10.5 m/s

Step-by-step 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.