Question

the displacement of free fall is directly proportional to ?

Answer 1

for free fall
$v = \sqrt{2gh} \\ {v}^{2} = 2gh \\ h = \frac{ {v}^{2} }{2g}$
so displacement is directly proportional to square of velocity

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Answer 2

We know, by Second Equation of Motion that-

s = ut+(at^2)/2

For a freely falling body, which starts from rest - u = 0

So, s = 0 +(at^2)/2

=> s = (at^2)/2

However, for a freely falling body, the acceleration is uniform as it is equal to the acceleration due to gravity(g) of the body.

So, here, a is constant

=>a/2 is constant

So, the equality sign disappears and the proportionality sign comes in its place.

So, s is directly proportional to t^2