A body falls freely from rest and covers the second half of…

A body falls freely from rest and covers the second half of its total height in $\Delta t = 1\ \mathrm{s}$. (a) What is its total fall time? (b) From what height was it released?

Explanation of Method

Second half of the height in one second.

Step 1: Two equations

Let the total time be $T$ and the height $H$. The first half of the height takes $T - 1$ seconds: $H = \tfrac12 g T^2$.

\[ \frac H2 = \tfrac12 g (T - 1)^2. \]

Step 2: (a) Total time

Divide the second equation by the first: \[\frac{(T - 1)^2}{T^2} = \frac12 \;\Longrightarrow\; \frac{T - 1}{T} = \frac{1}{\sqrt2} \;\Longrightarrow\; T = \frac{\sqrt2}{\sqrt2 - 1} = 2 + \sqrt2.\]

\[T = 2 + \sqrt2 \approx 3.41\ \mathrm{s}\]

Step 3: (b) Height

\[ H = \tfrac12 g T^2 = 4.905 \cdot 3.414^2. \]

\[H \approx 57.2\ \mathrm{m}\]

Check

First half: $ \tfrac12 \cdot 9.81 \cdot 2.414^2 \approx 28.6\ \mathrm{m} = H/2\ \checkmark $

$\text{distance fallen } (\mathrm{m})$ $t\ (\mathrm{s})$ $0$ $1$ $2.41$ $3.41$ $28.6$ $57.2$ $1\ \mathrm{s}$ $\text{halfway}$ $\text{ground}$ $\text{1st half: } 2.41\ \mathrm{s}$

Distance fallen against time. The first half of the height takes 2.41 s, the second half only 1 s, because the body is much faster by then.

Result

\[\boxed{\,T = 2 + \sqrt2 \approx 3.41\ \mathrm{s}\,}\]

\[\boxed{\,H \approx 57.2\ \mathrm{m}\,}\]

Remark

About the original solution. Its algebra reduced to "$2T - 1 = 2T - 1$", which is always true and cannot determine $T$; the value 3.4 s was then stated without a derivation. The ratio step above gives the exact $T = 2 + \sqrt2$, and with it $H \approx 57.2\ \mathrm{m}$ (rounding $T$ to 3.4 gives the slightly low 56.8 m).