Question #ee74d

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Question #ee74d

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Answer 1 · 2017-01-13T05:50:24

An object is released vertically downwards from a known height $h, units (feet)$ $h, "units (feet)"$ with initial velocity $u = 0, units (feet per second)$ $u=0, "units (feet per second)"$ at time $t = 0, units second)$ $t=0, "units second)"$ and allowed to fall freely under the force of gravity as shown in the figure below.

As it reaches ground ( $h = 0$ $h=0$ ), the time is noted. Let it be $t$ $t$ .

![http://keisan.casio.jp](https://useruploads.socratic.org/uqMHws1LRselWUAcEbTU_%25BC%25AB%25CD%25B3%25CD%25EE%25B2%25BCvh.gif)

General Kinematic equation equation governing the free fall under gravity is
$h = u t + \frac{1}{2} g t^{2}$ $h=ut+1/2g t^2$
Inserting given values and solving for $g$ $g$ acceleration due to gravity we get
$h = 0 \times t + \frac{1}{2} g t^{2}$ $h=0xxt+1/2g t^2$
$\Rightarrow g = \frac{2 h}{t^{2}}$ $=>g=(2h)/t^2$ , ${feet per second}^{2}$ $"feet per second"^2$

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Question #ee74d

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