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The distance travelled by a freely falling body is proportional to

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The mass of the body

The square of the acceleration due to gravity

The square of the time of fall

The time of fall

answer is C.

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Detailed Solution

The distance travelled by a freely falling body is proportional to the square of the time of fall.
We know that for a vertical motion,

s = ut + \frac{1}{2} g t^{2}

Where

s = distance

u = initial velocity

g = acceleration due to gravity

= 9.8 m / s^{2}

and

t = time

For a freely falling body, the initial velocity is zero, i.e.

u = 0

So the above equation becomes
⇒

s = 0 \times t + \frac{1}{2} g t^{2}

⇒

s = \frac{1}{2} g t^{2}

⇒

s \propto t^{2}

(since

g

is constant).

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