Question: In Figure, a solid ball rolls smoothly from rest (starting at height H = 6.0 m ) until it leaves the horizontal section at the end of the track, at height h = 2.0 m. How far horizontally from point Adoes the ball hit the floor?

H = 6.0m

h = 2.0 m

As the ball is rolling from a height (H) to another height (h), use the conservation of energy to calculate the speed of the ball. The next part of the motion of the ball would be projectile motion, so using kinematic equations find the horizontal distance.

Kinetic energy of the ball is given as-

${\mathbf{K}}_{\mathbf{f}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Mv}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}$

The ball’s potential energy is given as-

U = mgh

Form the law of conservation of energy,

$$\begin{gathered} mgH=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2+mgh \\ mgH=\frac{1}{2}m{r}^2{\omega }^2+\frac{2}{10}m{v}^2+mgh \\ mgH=m\times \left(\frac{7}{10}{v}^2+gh\right) \\ gH=\frac{7}{10}{v}^2+gh \end{gathered}$$

$$\begin{gathered} \frac{7}{10}{v}^2=g\times \left(H-h\right) \\ v=\sqrt{\frac{10}{7}\times g\times \left(H-h\right)} \\ v=\sqrt{\frac{10}{7}\times 9.8\times \left(6\text{m}-2\text{m}\right)} \\ v=7.48\text{m/s}\backslash \end{gathered}$$

As we know the height above the ground. So, using kinematic equation

$$\begin{gathered} \text{y}=-\frac{1}{2}{\text{gt}}^2 \\ 2\text{m}=-\frac{1}{2}\times 9.8m/s^2\times {\text{t}}^2 \\ \frac{4}{9.8m/s^2}=\text{t} \end{gathered}$$

$$\begin{gathered} \text{t}=\sqrt{0.41}\text{s} \\ t=0.64\text{s} \\ \text{Speed}=\frac{\text{distance}}{\text{time}} \\ d=7.48m/s \times 0.64\text{s} \\ =4.8\text{m} \end{gathered}$$

The horizontal distance traveled by the ball is 4.8 m