A body is dropped from a certain height at time $T = 0$ . It undergoes freefall and reaches the ground at time $T = t$ s. What is the expression for the height $h$ from which the body started its free fall?

Acceleration due to gravity is $g$ .

$h = g t^{2}$

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$h = g t$

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$h = \frac{1}{2} g t^{2}$

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$h = (\frac{1}{2}) g t^{- 2}$

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Solution

The correct option is C
$h = \frac{1}{2} g t^{2}$

The distance covered with initial velocity $u$ and undergiong constant accleration $a$ in time $t$ is given by:

$s = u t + \frac{1}{2} a t^{2}$

In this case the distance travelled is $h$ , acceleration is $g$

We know that $h = u t + (\frac{1}{2}) g t^{2}$ .

Since the body is initially at rest, $u = 0$ .

Therefore, $h = (\frac{1}{2}) g t^{2}$

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A body is dropped from a certain height at time T=0. It undergoes freefall and reaches the ground at time T=t s. What is the expression for the height h from which the body started its free fall? Acceleration due to gravity is g.

The correct option is C h=12gt2 The distance covered with initial velocity u and undergiong constant accleration a in time t is given by: s=ut+12at2 In this case the distance travelled is h, acceleration is g We know that h=ut+(12)gt2. Since the body is initially at rest, u=0. Therefore, h=(12)gt2

A body is dropped from a certain height at time $T = 0$ . It undergoes freefall and reaches the ground at time $T = t$ s. What is the expression for the height $h$ from which the body started its free fall?

Acceleration due to gravity is $g$ .