Question #3995c

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Question #3995c

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Answer 1 · 2017-05-27T17:34:31

$v \approx 9.81 t$ $v approx 9.81t$

Explanation:

(Here I am making an assumption that air resistance is negligible and that the height of the building is negligible compared to the radius of the Earth)

Using SUVAT equations,
$v = u + a t$ $v=u+at$
where $v$ $v$ is the velocity of the object as a function of time, $u$ $u$ is the initial velocity at $t = 0$ $t =0$ , $a$ $a$ is the acceleration of the object, and $t$ $t$ is any instant of time. As the object is dropped from a building, the horizontal component of its velocity will always remain 0, i.e. $v_{x} = 0$ $v_x=0$ , hence the vertical component will equal the velocity of the object, i.e. $v_{x} = v$ $v_x=v$ ; also, as it is dropped from a building, the vertical component of its initial velocity is 0, i.e. $u_{y} = 0$ $u_y=0$ .

As it is dropped from the building, the resultant force acting on the object is only the gravitational force acting on it due to Earth's gravitational field. The acceleration $a$ $a$ of the object is hence equal to acceleration due to gravity $g, \approx 9.81 m s^{- 2}$ $g, approx 9.81ms^-2$

Therefore, $v = v_{y} = u_{y} + g t \approx 0 + 9.81 t$ $v=v_y=u_y+ g t approx 0+ 9.81t$
$v \approx 9.81 t$ $v approx 9.81t$
Depending on the direction you choose, $v$ $v$ and $g$ $g$ can be positive or negative

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Question #3995c

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