(1 point) NASA launches a rocket at $t=0$ seconds. Its height, in meters above sea-level, as a function of time is given by $h(t)=-4.9 t^{2}+277 t+353$. Assuming that the rocket will splash down into the ocean, after how many seconds does splashdown occur? Round your answer to the nearest hundredth. The rocket splashes down after seconds. How high above sea-level does the rocket get at its peak? Round your answer to the nearest hundredth. The rocket peaks at meters above sea-level.

Step 1 ：The rocket splashes down when its height is zero, i.e., when \(h(t) = 0\). This is a quadratic equation, and we can solve it by using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this case, \(a = -4.9\), \(b = 277\), and \(c = 353\). We are looking for the positive root of this equation, as time cannot be negative.

Step 2 ：By substituting the values into the quadratic formula, we get two solutions for \(t\), \(t1 = -1.2468667308378916\) and \(t2 = 57.77747897573585\). Since time cannot be negative, the rocket splashes down after approximately \(t2 = 57.77747897573585\) seconds.

Step 3 ：The height of the rocket at its peak is the maximum value of the function \(h(t)\). This occurs at the vertex of the parabola, which is given by \(t = -\frac{b}{2a}\). We can substitute this value back into the function to find the maximum height.

Step 4 ：By substituting the values into the formula for the vertex of a parabola, we get \(peak\_time = 28.265306122448976\). Substituting this value back into the function \(h(t)\), we get the maximum height of the rocket, \(peak\_height = 4267.744897959183\).

Step 5 ：Final Answer: The rocket splashes down after approximately \(\boxed{57.78}\) seconds. The rocket peaks at approximately \(\boxed{4267.74}\) meters above sea-level.