From a seaside cliff of height H = 80 m , a body is launched…
From a seaside cliff of height $H = 80\ \mathrm{m}$, a body is launched at $v_0 = 400\ \mathrm{m/s}$ and $\alpha = 45^\circ$ to the horizontal. Where does it hit the water, and how long is it in the air (no air resistance)?
Explanation of Method
A shot at 45° from a seaside cliff.
Step 1: Components
\[ v_{0x} = v_{0y} = 400 \cdot \tfrac{\sqrt2}{2} \approx 282.84\ \mathrm{m/s}. \]
Step 2: Flight time
With the water at $y = 0$: $y = H + v_{0y}t - \tfrac12 g t^2 = 0$, i.e. $4.905\,t^2 - 282.84\,t - 80 = 0$: $t = \frac{282.84 + \sqrt{282.84^2 + 4 \cdot 4.905 \cdot 80}}{9.81}$.
\[ t = \frac{282.84 + 285.61}{9.81}. \]
\[t \approx 57.9\ \mathrm{s}\]
Step 3: Landing point
\[ x = v_{0x}t = 282.84 \cdot 57.95. \]
\[x \approx 16\,390\ \mathrm{m} \approx 16.4\ \mathrm{km} \ \text{from the foot of the cliff}\]
Result
\[\boxed{\,t \approx 57.9\ \mathrm{s}\,}\]
\[\boxed{\,x \approx 16\,390\ \mathrm{m} \approx 16.4\ \mathrm{km} \ \text{from the foot of the cliff}\,}\]
Remark
About the original solution. The method is right, but the root was evaluated as 58.2 s instead of 57.95 s, which made the distance 16 470 m instead of about 16 390 m. From level ground the flight would last 57.66 s; the 80 m cliff adds only 0.28 s.