Question

A physics book of unknown mass is dropped 4.5 m.A) write an equation and find the speed the book has right before it hits the ground.

Answer 1

Given,

The height from which the book was dropped, h=4.5 m

When the book is at the height h, at rest it possesses only the potential energy. As it falls, it loses its potential energy and gains kinetic energy. Just before the book hits the ground, the book loses all its potential energy and possesses only kinetic energy.

Thus the potential energy of the book when it was at the height h is equal to its kinetic energy just before it hits the ground.

That is,

`\begin{gathered} (1)/(2)mv^2=\text{mgh} \\ \Rightarrow(v^2)/(2)=gh \\ \Rightarrow v=\sqrt[]{2gh} \end{gathered}`

Where v is the speed of the book just before it hits the ground and g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} v=\sqrt[]{2*9.8*4.5} \\ =9.39\text{ m/s} \end{gathered}`

Thus the speed of the book right before it hits the ground is 9.39 m/s.