Question

An apple falls from a tree. it hits the ground at a speed of about 8 m/s . what is the approximate height of the tree?

Answer 1

V=sqrt(2gh)
8m/s=sqrt(2*9.8m/s^2*h)
h=3.3m

Answer 2

The motion of the apple is a free fall motion, so it is a uniformly accelerated motion. Therefore, we can use the following equation:

$v^2-u^2 =2gh$

where

v=8 m/s is the speed of the apple when it hits the ground

u=0 is the initial speed of the apple

$g=9.81 m/s^2$ is the free-fall acceleration

h is the height from which the apple has fallen

Substituting data and rearranging the equation, we can find the height of the tree, h:

$h=\frac{v^2}{2g}=\frac{(8 m/s)^2}{2(9.81 m/s^2)}=3.26 m$