NASA launches a rocket at $t=0$ seconds. Its height, in meters above sea-level, as a function of time is given by $h(t)=-4.9 t^{2}+310 t+157$.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after seconds.
How high above sea-level does the rocket get at its peak?
The rocket peaks at meters above sea-level.

Step 1 ：Given the equation for the height of the rocket as a function of time, \(h(t) = -4.9 t^{2}+310 t+157\).

Step 2 ：The rocket splashes down when \(h(t) = 0\). Solving the equation \(-4.9 t^{2}+310 t+157 = 0\) for \(t\), we get two solutions: \(t = -0.502461004518698\) and \(t = 63.7677671269677\). Since time cannot be negative, the rocket splashes down approximately 63.77 seconds after launch.

Step 3 ：The maximum height of the rocket occurs at the vertex of the parabola represented by the equation. The \(t\)-coordinate of the vertex of a parabola given by \(h(t) = at^{2} + bt + c\) is given by \(-\frac{b}{2a}\). Substituting \(a = -4.9\) and \(b = 310\) into this formula, we get \(t = 31.632653061224488\).

Step 4 ：Substituting this value of \(t\) back into the equation \(h(t) = -4.9 t^{2}+310 t+157\), we find that the maximum height of the rocket is approximately 5060.06 meters above sea-level.

Step 5 ：\(\boxed{\text{Final Answer: The rocket splashes down approximately 63.77 seconds after launch. The rocket reaches a peak height of approximately 5060.06 meters above sea-level.}}\)