# 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?

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

${v}^{2} - {u}^{2} = 2 g h$

Inserting given values in SI units we get

${\left(4.38\right)}^{2} - {\left(0\right)}^{2} = 2 \times 9.81 \times h$
$\implies h = {\left(4.38\right)}^{2} / \left(2 \times 9.81\right) = 97.7 \setminus c m$

Height given in $c m$ as speed given in $c m {s}^{-} 1$

Feb 24, 2018

Alternate solution.

#### Explanation:

Let height of the window be $= h \setminus m$ above the ground.
Let the book be at a height ${h}_{1} \setminus m$ above ground when the desired speed is achieved.
Let the student drop the book with initial velocity $= 0$

${v}^{2} - {u}^{2} = 2 g s$
${\left(4.38\right)}^{2} - {\left(0\right)}^{2} = 2 \times 9.81 \times \left(h - {h}_{1}\right)$
$\implies \left(h - {h}_{1}\right) = {\left(4.38\right)}^{2} / \left(2 \times 9.81\right)$
$\implies \left(h - {h}_{1}\right) = 0.98$
$\implies {h}_{1} = 0.98 - h \setminus m$