You attach a meter stick to an oak tree, such that the top of the meter stick is 2.87 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.296 seconds to pass the length of the meter stick, how high ℎ0 above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down?

3.06 m

Explanation:

Given:

y₀ = h

y₁ = 2.87 m

y₂ = 1.87 m

v₀ = 0 m/s

a = -9.8 m/s²

t₂ − t₁ = 0.296 s

y₁ = y₀ + v₀ t₁ + ½ at₁²

2.87 = h − 4.9t₁²

y₂ = y₀ + v₀ t₂ + ½ at₂²

1.87 = h − 4.9t₂²

Subtracting the equations:

1 = (h − 4.9t₁²) − (h − 4.9t₂²)

1 = h − 4.9t₁² − h + 4.9t₂²

1 = 4.9 (t₂² − t₁²)

1 = 4.9 (t₂ − t₁) (t₂ + t₁)

1 = 4.9 (0.296) (0.296 + 2t₁)

t₁ = 0.197

Solve for h:

2.87 = h − 4.9 (0.197)²

h = 3.06

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