Question

A 33.0-kg child starting from rest slides down a water slide with a vertical height of 15.0 m. (Neglect friction.)

Answer 1

This question involves the concept of the conservation of energy.

(a) The child's speed halfway will be "12.13 m/s".

(b) The child's speed three-fourth way will be "14.86 m/s".

(a)

Using the law of conservation of energy:

Potential Energy Lost = Kinetic Energy Gained

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\v = \sqrt{2gh}$

where,

v = speed = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = halfway = height/2 = 15 m/2 = 7.5 m

Therefore,

$v = \sqrt{(2)(9.81\ m/s^2)(7.5\ m)}$

v = 12.13 m/s

(b)

Using the law of conservation of energy:

Potential Energy Lost = Kinetic Energy Gained

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\v = \sqrt{2gh}$

where,

v = speed = ?

g = acceleration due to gravity = 9.81 m/s²

h = height lost = three-fourth way down = $\frac{3}{4}height = \frac{3}{4}(15\ m) = 11.25\ m$

Therefore,

$v = \sqrt{(2)(9.81\ m/s^2)(11.25\ m)}$

v = 14.86 m/s

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture shows the law of conservation of energy.