Question

My question involves the Derivatives of Rates of Change:

The question revolves around 4 parts around the rates of change for a rocket on the moon.
the formula is
s = 24t - 0.8t^2
*s is the height in meters
*t is time in seconds

It's divided into 4 parts, and while I got 3 of them, the last one has me stumped.

It asks for the time it takes for the rocket to reach half of it's max height.

I found the time it takes to reach max height to be 15 seconds, and max height to be 180.

How can I solve this algebraically? I know it's around 4.39 secs.

Answer 1

since you have the max height at \(180\) ( i didn't check it, just assume you are right) then half the max height is \(90\) set
\[24t-0.8t^2=90\] and solve for \(t\)

Answer 2

^I'll expand on that right now for you.
v = 24 - 1.6t = s'
a = - 1.6

I found t with the velocity function, and plugged it into the original function and max height

Answer 3

^for max height

Answer 4

I end up having square root a negative, which isn't possible?