Question

A toy rocket launched vertically into the air has position function s(t) = 64√t − 8t^2, where s = 0 is the ground, s is measured in meters and t > 0 is measured in seconds.

How many seconds does the rocket spend traveling upward?

Answer 1

and After how many seconds does the rocket return to the ground?

Answer 2

The rocket travels upwards until s is the highest. Then it starts to fall down. So find the maximum point for the function. I think.

When the rocket returns to the ground, s is zero again. so find that.

Answer 3

could you work it out?

Answer 4

at the maximum, the slope of the function is zero. Because at that point the tangent line is simply horizontal. the slope is zero, so the derivative at that point is zero.

Find the derivative of the function. Then make it = 0
Try it.

Answer 5

t=4?

Answer 6

that is when the function equals zero. Which is when the rocket returns to the ground. You have the second question.

For the first question, you need to find the value of t when the DERIVATIVE is zero.

Answer 7

is the derivative \[64\times 1\div2T ^{-1/2}-16t ?\]

Answer 8

it can be simplified but yeah, that is correct.

Answer 9

simplifies it would be 32/(sqrt t)-16t?

Answer 10

technically your first one may not be right, because of order of operations and stuff, but I think you got it right. The second one is correct, but simplified again it is:

\[32t^{-1/2}\]

Answer 11

ok when you set this equal to 0 you get 32^2 or 1024?

Answer 12

I meant
\[32t^{-1/2}-16t\]

But, no, you seem to have miscalculated. maybe show your working?

Answer 13

i got that much and then you solve for t correct and i get 4? and thats not the answer so Im confused

Answer 14

i got that much and then you solve for t correct and i get 4? and thats not the answer so Im confused

Answer 15

All you have to do now is solve this:
\[32t^{-\frac{1}{2}}-16t=0\]
Almost there. Just try again.