Question

A student drops his textbook out of a window in a tall building. How high above the ground is the book when it is falling at a speed of 438 cm/s?

Answer 1

It is presumed that the given falling speed of book just is as it hits the ground.

Explanation:

Applicable kinematic expression is

`v^2-u^2=2gh`

Inserting given values in SI units we get

`(4.38)^2-(0)^2=2xx9.81xxh`
`=>h=(4.38)^2/(2xx9.81)=97.7\ cm`

Height given in `cm` as speed given in `cms^-1`

Answer 2

Alternate solution.

Explanation:

Let height of the window be `=h\ m` above the ground.
Let the book be at a height `h_1\ m` above ground when the desired speed is achieved.
Let the student drop the book with initial velocity `=0`

For free fall under gravity applicable kinematic expression is

`v^2-u^2=2gs`

Inserting given values in SI units we get

`(4.38)^2-(0)^2=2xx9.81xx(h-h_1)`
`=>(h-h_1)=(4.38)^2/(2xx9.81)`
`=>(h-h_1)=0.98`
`=>h_1=0.98-h\ m`