A pendulum, swinging back and forth, rises at the end of its swing to a position $15cm$. higher than its lowest point. How fast is it going at the lowest point?

Step 1: Given that

Given that:-

Maximum height $h=$$15cm$

$\Rightarrow h=15cm=0.15m$

Step 2: Finding the velocity using the law of conservation of energy,

The potential energy of the pendulum at the maximum height is equal to the kinetic energy of the pendulum at the lowest point,

$$\begin{gathered} \Rightarrow mgh=\frac{1}{2}m{v}^2 \\ \Rightarrow 9.8\times 0.15=\frac{1}{2}{v}^2 \\ \Rightarrow 9.8\times 0.15\times 2=v^2 \\ \Rightarrow v=\sqrt{2.94} \\ \Rightarrow v=1.7\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence, pendulum $1.7\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ is the velocity of the pendulum at the lowest point.