Question

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.141 seconds to pass the length of the meter stick, how high 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?

Answer 1

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

$u = 0m/s$

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

$s = ut+\frac{1}{2} at^2$

Replacing,

$1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2$

$u =6.4013m/s$

The height of the acorn above the meter stick can be calculated as,

$v^2 = u^2 +2gh$

$h = \frac{v^2-u^2}{2g}$

$h = \frac{6.4013^2-0^2}{2(9.8)}$

$h = 2.0906m$

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

$h = 2.0906+1.87$

$h = 3.9606m$