A body falls freely from rest. In the last Delta t = 0.6 s…
A body falls freely from rest. In the last $\Delta t = 0.6\ \mathrm{s}$ of its fall it covers $\Delta s = 30\ \mathrm{m}$. From what height was it dropped?
Explanation of Method
Height from the distance covered in the last 0.6 s.
Step 1: Speed in the middle of the interval
In uniformly accelerated motion, the average speed over an interval equals the speed at its middle: \[v_{\text{mid}} = \frac{\Delta s}{\Delta t} = \frac{30}{0.6} = 50\ \mathrm{m/s}.\]
The body reaches 50 m/s at $t_{\text{mid}} = 50/9.81 \approx 5.097\ \mathrm{s}$.
Step 2: Total fall time
The interval ends 0.3 s after its middle, at the ground: $T = t_{\text{mid}} + \tfrac{\Delta t}{2} = 5.097 + 0.3 \approx 5.40\ \mathrm{s}$.
Step 3: Height
\[ h = \tfrac12 g T^2 = 4.905 \cdot 5.397^2. \]
\[h \approx 143\ \mathrm{m}\]
Check with the original equation
$ 30 = \tfrac{9.81}{2}\,(1.2\,t_1 + 0.36) \Rightarrow 1.2\,t_1 = 6.116 - 0.36 \Rightarrow t_1 \approx 4.80\ \mathrm{s} $, so $ T = 5.40\ \mathrm{s}\ \checkmark $
$ \tfrac12 g (5.397^2 - 4.797^2) = 142.9 - 112.9 = 30.0\ \mathrm{m}\ \checkmark $
Result
\[\boxed{\,h \approx 143\ \mathrm{m}\,}\]
Remark
About the original solution. The equation was right, but when solving for $t_1$ it did not divide both terms by 1.2, getting 5.56 s instead of 4.80 s. That made the height 186 m instead of about 143 m.