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?

2 Answers
Feb 17, 2018

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#

Feb 24, 2018

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#