From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is _______ m. (Take $$g = 10$$ m/s$$^2$$ and neglect the air resistance)

Correct Answer: 13

For a freely falling body:

v² = 2g(h₀ − h)

Here initial height h₀ = 18 m

So:

v² = 2g(18 − h)

Given condition:

|v| = g

So:

v² = g²

equate:

g² = 2g(18 − h)

divide by g:

g = 2(18 − h)

10 = 36 − 2h

2h = 26

h = 13 m

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