Can somebody please help me
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 + 217t + 185.

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?

The rocket splashes down afterseconds.

How high above sea-level does the rocket get at its peak?

The rocket peaks atmeters above sea-level

a) 45.1 seconds

b) 2587.5 ft

Step-by-step explanation:

a)

The height of the rocket as a function of time is given by:

where

t is the time (in seconds)

-4.9 represents the acceleration of gravity (in ft/s^2)

+217 represents the initial velocity (in ft/s)

+185 represents the initial heigth of the rocket at t = 0 (in feet)

Here we want to find the time t at which the rocket splashes down, so the time t at which the height of the rocket is zero:

h(t) = 0

To do so, we substituting h(t) = 0 into the expression:

Using the formula to find t:

which gives two solutions:

Neglecting the negative solution, the rocket splashes down after 45.1 seconds.

b)

In order to find the maximum height of the rocket, we have to find the time t at which the derivative of the function h(t) becomes zero.

The function is

Therefore, its derivative vs t is:

This expression represents the velocity of the rocket at time t.

The derivative becomes zero when h'(t) = 0, so:

So, the rocket reaches the maximum height after 22.1 seconds. Substituting this value into h(t), we find the maximum height  reached by the rocket:

answer is 67%

step-by-step explanation: