What height does a frictionless playground slide need so that a 35 kg child reaches the bottom at a speed of 4.5 m/s?

If there is no external force, the sum of any system's initial kinetic and potential energy equals the sum of the system's ultimate kinetic and potential energy. It denotes that the total of kinetic and potential energy changes is zero.

$\Delta K+\Delta U=0$

we have $\Delta K$is the difference between kinetic energy and potential energy $\Delta U$is the difference between potential energy and actual energy.

By Diagram:

The sum of changes in kinetic and potential energy is 0 since no external effort is done on the system. Therefore,

$\Delta K+\Delta U=0$

We have $\Delta K$is the boy's kinetic energy changing and $\Delta U$is the boy's potential energy changing.

The following is the equation for kinetic energy change:

$\Delta K=\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_0^2$

we see$v_0$is the young boy initial speed,$v_1$is the boy's final speed as he reaches the ground

The following is the equation for the change in potential energy:

$\Delta U=mg{y}_1-mg{y}_0$

$=mg\left(y_1-y_0\right)$

we have $y_0$is the slide's height and$y_1$is the ground's elevation.

From,

$\Delta K+\Delta U=0$

as from above,

$\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_0^2+mg\left(y_1-y_0\right)=0$

$v_0^2-v_1^2=-2g\left(y_0-y_1\right)$

$\left(y_0-y_1\right)=\frac{v_0^2-v_1^2}{-2g}$

Putting $0\mathrm{m}/\mathrm{s}$for $v_0,4.5\mathrm{m}/\mathrm{s}$for $v_1,9.8\mathrm{m}/{\mathrm{s}}^2$for $g$and $0\mathrm{m}$for $y_1$.

$y_0-0\mathrm{m}=\frac{(0\mathrm{m}/\mathrm{s}{)}^2-(4.5\mathrm{m}/\mathrm{s}{)}^2}{-2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}$

$y_0=1\mathrm{m}$

As a result, the frictionless slide's height must be$1\mathrm{m}$.