Question

A boy uses a rope attached to the top of a ceiling to swing to and fro .The lowest and highest point of the swing is 1.5 and 2.5 above the ground respectively .What is the boy's max speed

Answer 1

From the conservation of energy:

$$0.5mv^2=mg(H-h)$$

Thus, maximum speed is

$$v=\sqrt{g(H-h)}=\sqrt{9.8(2.5-1.5)}=3.1\frac{m}{s}$$