If a body is dropped from a height of $ h $, at what velocity will it touch the ground?

Question

A body starts from rest and moves with constant acceleration. If it covers a distance of $ 40 \, \text{m} $ in 4 seconds, then its acceleration is:

Answer 1

Step 1: By energy conservation, $ mgh = \tfrac12 mv^2 $.
Step 2: The mass $m$ cancels: $ v^2 = 2gh $.
\[ \boxed{v = \sqrt{2gh}} \]

Answer 2

A third way is dimensional analysis combined with a limiting-case check: since $v$ must depend only on $g$ (dimensions $LT^{-2}$) and $h$ (dimension $L$), the only combination with dimensions of velocity ($LT^{-1}$) is proportional to $\sqrt{gh}$; a constant of proportionality must then be fixed by comparing to a known simple case.

Dimensionally, $ v = k\sqrt{gh} $ for some dimensionless constant $k$. Testing against the well-known result for a fall from a small height (verified independently by direct integration of $v=\int g\,dt$ alongside $h=\int v\,dt$, which gives $v^2=2gh$) fixes $k=\sqrt2$, so $ v=\sqrt{2gh} $.

$ \sqrt{gh} $: This corresponds to $k=1$, which does not match the fixed constant $k=\sqrt2$; incorrect.
$ \sqrt{2gh} $: This corresponds to $k=\sqrt2$, matching the fixed constant found above.
$ 2\sqrt{gh} $: This corresponds to $k=2$, which is too large compared to the fixed constant $\sqrt2$; incorrect.
depends on mass of the body: Dimensional analysis shows mass does not even appear in the relevant quantities $g$ and $h$ that determine $v$, so this option is incorrect.

Fixing the dimensionless constant confirms the same formula as the energy and kinematic derivations.

Therefore, the correct answer is $ \sqrt{2gh} $.

Answer 3

A third way is dimensional analysis combined with a limiting-case check: since $v$ must depend only on $g$ (dimensions $LT^{-2}$) and $h$ (dimension $L$), the only combination with dimensions of velocity ($LT^{-1}$) is proportional to $\sqrt{gh}$; a constant of proportionality must then be fixed by comparing to a known simple case.

Dimensionally, $ v = k\sqrt{gh} $ for some dimensionless constant $k$. Testing against the well-known result for a fall from a small height (verified independently by direct integration of $v=\int g\,dt$ alongside $h=\int v\,dt$, which gives $v^2=2gh$) fixes $k=\sqrt2$, so $ v=\sqrt{2gh} $.

$ \sqrt{gh} $: This corresponds to $k=1$, which does not match the fixed constant $k=\sqrt2$; incorrect.
$ \sqrt{2gh} $: This corresponds to $k=\sqrt2$, matching the fixed constant found above.
$ 2\sqrt{gh} $: This corresponds to $k=2$, which is too large compared to the fixed constant $\sqrt2$; incorrect.
depends on mass of the body: Dimensional analysis shows mass does not even appear in the relevant quantities $g$ and $h$ that determine $v$, so this option is incorrect.

Fixing the dimensionless constant confirms the same formula as the energy and kinematic derivations.

Therefore, the correct answer is $ \sqrt{2gh} $.

If a body is dropped from a height of \( h \), at what velocity will it touch the ground?

Show Hint

The Correct Option is B

Approach Solution - 1

Approach Solution -2

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