Question #35de3

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Question #35de3

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Answer 1 · 2016-09-26T17:20:58

$2.71 s$ $2.71s$ , rounded to two decimal places.

Explanation:

Free fall under gravity is governed by the kinematic equation
$h = u t + \frac{1}{2} g t^{2}$ $h=ut+1/2g t^2$
Where $h$ $h$ is height from which body starts falling,
$u$ $u$ is its initial velocity
$g$ $g$ is acceleration due to gravity and is $9.81 m s^{- 2}$ $9.81ms^-2$
and $t$ $t$ is the time taken for the fall to reach ground level.

In this question we take ground level as the sea level. Inserting given quantities we get
$36 = 0 \times t + \frac{1}{2} \times 9.81 \times t^{2}$ $36=0xxt+1/2xx9.81xx t^2$
$\Rightarrow \frac{1}{2} \times 9.81 \times t^{2} = 36$ $=>1/2xx9.81xx t^2=36$
$\Rightarrow t^{2} = \frac{36 \times 2}{9.81}$ $=> t^2=(36xx2)/9.81$
$\Rightarrow t = \sqrt{\frac{36 \times 2}{9.81}}$ $=> t=sqrt((36xx2)/9.81)$
$\Rightarrow t = \pm 2.71$ $=>t=+-2.71$ , rounded to two decimal places.
ignoring the $- v e$ $-ve$ root as time can not be negative
$t = 2.71 s$ $t=2.71s$

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Question #35de3

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