Question

At what time does splashdown occur ?how high above sea-level does the rocket get at its peak ?

Answer 1

Given the function:

`h(t)=-4.9t^2+136t+311`

Given that NASA launches a rocket at t= 0 seconds.

Where h is the height in meters above sea level.

Let's solve for the following:

• (a). The time the rocket splashes the water.

When the rocket splashes the water, the height above sea-level will be 0 meters.

To solve for t, set h(t) for 0 and solve.

We have:

`-4.9t^2+136t+311=0`

Now, let's solve using the quadratic formula:

`x=(-b\pm√(b^2-4ac))/(2a)`

Where:

a = -4.9

b = 136

c = 311

Plug in the values into the formula and solve for t:

`\begin{gathered} t=(-136\pm√(136^2-4(-4.9)(311)))/(2(-4.9)) \\ \\ t=(-136\pm√(18496-(-6095.6)))/(-9.8) \\ \\ t=(-136\pm√(18496+6095.6))/(-9.8) \\ \\ t=(-136\pm√(24591.6))/(-9.8) \\ \\ t=(-136\pm156.82)/(-9.8) \\ \\ \end{gathered}`

Solving further:

`\begin{gathered} t=(-136-156.82)/(-9.8),(-136+156.82)/(-9.8) \\ \\ t=(−292.82)/(-9.8),(20.82)/(-9.8) \\ \\ t=29.88,-2.12 \end{gathered}`

Thus, we have the solutions:

t = 29.88 and t = -2.12

Since the time cannot be negative, let's take the positive solution.

Therefore, the rocket splashes down after 29.88 seconds.

• (b). Let's find the peak.

To find the maximum time, apply the fomula:

`t=-(b)/(2a)`

Thus, we have:

`\begin{gathered} t=-(136)/(2(-4.9)) \\ \\ t=-(136)/(-9.8) \\ \\ t=(136)/(9.8) \\ \\ t=13.88 \end{gathered}`

The rocket gets to its peak at 13.88 seconds.

Now, to find the height at that time, substitute 13.88 for t in h(t) and solve for h(13.88):

`\begin{gathered} h(13.88)=-4.9(13.88)^2+136(13.88)+311 \\ \\ h(13.88)=-944.01+1887.35+311 \\ \\ h(13.88)=1254.67\text{ m} \end{gathered}`

Therefore, the rocket peaks at 1254.67 above seal level.

ANSWER:

• 29.88 seconds

• 1254.67 meters