A rocket starts vertically upward with constant acceleration…
A rocket starts vertically upward with constant acceleration $a_1 = 4g$. After $t_0 = 8\ \mathrm{s}$ a part of it separates and is pushed sideways with speed $v_0 = 80\ \mathrm{m/s}$. Where does this part land? Ignore air resistance.
Explanation of Method
Where a part separated from a rising rocket lands.
Step 1: The rocket at separation
\[ h_0 = \tfrac12 a_1 t_0^2 = \tfrac12 \cdot 39.24 \cdot 64 \approx 1255.7\ \mathrm{m}, \]
\[ v_y = a_1 t_0 = 39.24 \cdot 8 = 313.9\ \mathrm{m/s}. \]
Step 2: The part's initial velocity
It keeps the rocket's upward speed and gets 80 m/s sideways; from now on only gravity acts: $v_{x0} = 80\ \mathrm{m/s}$.
\[ v_{y0} = 313.9\ \mathrm{m/s}. \]
Step 3: Time until it lands
\[ h_0 + v_{y0}t - \tfrac12 g t^2 = 0 \;\Longrightarrow\; t = \frac{v_{y0} + \sqrt{v_{y0}^2 + 2gh_0}}{g}, \]
\[ t = \frac{313.9 + 351.0}{9.81}. \]
\[t \approx 67.8\ \mathrm{s}\]
Step 4: Landing point
\[ x = v_{x0}\,t = 80 \cdot 67.78. \]
\[x \approx 5420\ \mathrm{m} \approx 5.4\ \mathrm{km} \ \text{from the launch pad}\]
Check
It rises another $ \tfrac{313.9^2}{2 \cdot 9.81} \approx 5023\ \mathrm{m} $ in 32.0 s, to 6278 m; falling 6278 m takes $ \sqrt{2 \cdot 6278/9.81} \approx 35.8\ \mathrm{s} $; total $ 67.8\ \mathrm{s}\ \checkmark $
Result
\[\boxed{\,t \approx 67.8\ \mathrm{s}\,}\]
\[\boxed{\,x \approx 5420\ \mathrm{m} \approx 5.4\ \mathrm{km} \ \text{from the launch pad}\,}\]
Remark
About the original solution. Its formula $t = \tfrac{v_{y0}}{g} + \sqrt{\tfrac{2h_0}{g} + \left(\tfrac{v_{y0}}{g}\right)^2}$ is right, but it evaluates to 32.0 + 35.8 = 67.8 s, not 32.01 s (only the first term was kept). So the part lands about 5.4 km away, not 2.56 km. (Also $h_0 = 1255.7\ \mathrm{m}$, not 1256.64.)