What will be the magnitude of the velocity of the child at the bottom of the slide no. 1? [Pg. No. 116]

Potential energy of the child at the top of the slide = $mgh$

Let the magnitude of velocity of the child on reaching the bottom of the slide be $v$. Then, the kinetic energy of the child at the bottom of the

$$\text{slide} = \frac{1}{2}mv^2$$

The potential energy of the child at the top of the slide gets converted entirely to the child's kinetic energy at the bottom of the slide (neglecting the effects of friction). Thus, equating the two, we obtain

$$\frac{1}{2}mv^2 = mgh$$

$$v = \sqrt{2gh}$$

As this velocity of the child at the bottom of the slide depends only upon the height $h$, the shape of the slide or the mass of the child should not matter.