NASA launches a rocket at $t=0$ seconds. Its height, in meters above sea-level, as a function of time is given by $h(t)=-4.9 t^{2}+196 t+262$
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after seconds.

Step 1 ：The rocket splashes down when its height is zero, i.e., when \(h(t) = 0\). So, we need to solve the equation \(-4.9 t^{2}+196 t+262 = 0\) for \(t\). This is a quadratic equation, and we can solve it using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -4.9\), \(b = 196\), and \(c = 262\). However, since time cannot be negative, we only consider the positive root.

Step 2 ：The result is negative, which doesn't make sense in the context of this problem since time cannot be negative. This means we should consider the other root of the quadratic equation, which is given by \(t = \frac{-b - \sqrt{b^2 - 4ac}}{2a}\).

Step 3 ：Final Answer: The rocket splashes down after \(\boxed{41.29}\) seconds.