Question

A ripe pear falls from a tree and lands on the ground. As it hit the ground, the pear had a speed of 34 feet per second. Approximately how far up in the tree was the pear before it fell? Use the formula below, and assume that the acceleration due to gravity is 32 feet per second squared: s=pi2gh

Answer 1

The potential energy is converted into kinetic energy.
 Therefore, for conservation of energy we have:
 mgh = (1/2) m * v ^ 2
 From here we clear the speed:
 v = root (2 * g * h)
 The height will then be:
 h = v ^ 2 / (2 * g)
 Substituting the values:
 h = (34) ^ 2 / (2 * (32))
 h = 18.0625 feet
 Answer: The pear before it fell was 18.0625 feet up in the tree.

Answer 2

Answer:

18.0625 feet

Step-by-step explanation:

Given : The pear had a speed of 34 feet per second.

To Find: Approximately how far up in the tree was the pear before it fell?

Solution:

We are given that assume that the acceleration due to gravity is 32 feet per second squared

Formula = $s=\sqrt{2gh}$

Where s is the speed , g is the acceleration due to gravity and h is the height

So, $34=\sqrt{2(32)h}$

$34^2=64h$

$1156=64h$

$\frac{1156}{64}=h$

$18.0625=h$

Hence the pear was at height of 18.0625 feet before it fell.