Questions: 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+286 t+304 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.

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+286 t+304

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.

Transcript text: NASA launches a rocket at $t=0$ seconds. Its height, in meters above sea-level, as a function of time is given by $h(t)=-4.9 t^{2}+286 t+304$ Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? The rocket splashes down after $\square$ seconds. How high above sea-level does the rocket get at its peak? The rocket peaks at $\square$ meters above sea-level.

Solution

Solution Steps

Step 1: Finding the Splashdown Time

To find the splashdown time, we solve the equation $at^2 + bt + c = 0$ for $t$ using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Considering only the positive root for a real and positive time value, we get the splashdown time as $t = -1.04$ seconds.

Step 2: Finding the Peak Height

The peak height is found by determining the vertex of the parabola. The time at which the peak occurs is $t_{\text{peak}} = \frac{-b}{2a} = 29.184$. Substituting $t_{\text{peak}}$ back into the original equation gives the maximum height, $h_{\text{max}} = a(t_{\text{peak}})^2 + b(t_{\text{peak}}) + c = 4477.27$ meters.