Question

A child slides down a frictionless slide from a height of 1.8 m. How fast will she be going at the bottom of the slide ? (g=9.8 $m{s}^{-2}$)

Answer 1

The child is moving at the bottom of the slide. We're going to do this using conservation of energy.

At the top of the slide, the child has all gravitational potential energy, given by

U_i = mgh

where m is the mass of the child

$g=9.8$

the height of the slide h = L sin 45^o

At the bottom of the slide, all of that energy will have been converted to kinetic energy:

K_f = ½ M V ²

we aren't losing any energy to friction, we must have

L sin 45^o = ½ M V ²

$L=3.2$

Solving $v=6.7m/s$