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 ) = − 4.9t^2 + 202t + 318 . assuming that the rocket will splash down into the ocean, at what time does splashdown occur? the rocket splashes down . how high above sea-level does the rocket get at its peak? the rocket peaks above sea-level

The rocket splashes down after 43 seconds.

The rocket peaks at 2400. meters above sea-level.

Step 1

We realise that the rocket will hit the ground when our function , i.e

Step 2

Use the quadratic formula to solve for the time. For the general quadratic equation , the general solution to the equation is

In our case, the quadratic equation has a solution

Step 3

In this step we pick the value of that makes physical sense. Since the NASA team launced the rocket at , the rocket reaches the bottom when . This can be rounded up to seconds.

Step 1

The first step is to realist that since the coefficient of the quadratic term is negative, the graph of this function will face downwards. From this we can gather that the maximum value of this function occurs at the vertex.

Step 2

In this step we calculate the t coordinate of the of the vertex of this function. For any quadratic equation , the vertex occurs when ,

For the equation , the vertex occurs when ,

Step 3

We substitute the value of from step 2 above into to get the maximum value of this function.  This calculation is shown below,

This can be rounded up to 2400m.

8

step-by-step explanation:

this is a subtraction problem.

10 - 2 = 8

66.666666667%

step-by-step explanation:

divide 150 by 100

divide 100 by 1.5

there you go!

decreased by 33.333333334%