Question

A branch falls from the top of a 95.0 m tall redwood tree, starting from rest. How fast is it moving when it reaches the ground?

Answer 1

$v = \sqrt{2gh}$
$v = \sqrt{20 \times 95}$
$v = 43.5$
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Answer 2

We can solve this problem by using linear motion equations.

Let U = initial velocity of branch = 0

      a = acceleration due to gravity = 10m /s²

     S = traveled distance = 95 m

     V = velocity when it reaches the ground

Using

$V^{2} = U^{2} + 2aS\\ V^{2} = 0 + 2 *10*95\\ V^{2} = 1900$

        V = 43.6 m/s

Answer : Velocity = 43.6 m/s