Question

A small rocket is shot into the air. Its height (in feet) after t seconds is given by the function: h(t)=-16t^2+15t+48. Find the maximum height of the skyrocket.

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

Have you taken calculus?

Answer 3

No. I'm in algebra 1. I have a final exam tomorrow and I forgot how to do this. This question is on my study guide. Help!!

Answer 4

???

Answer 5

OK. I just asked to know what method would be more suitable for you. Anyway, you need to write \(h(t)\) in a form of a complete square, that's \(h(t)= -16(t-\frac{15}{32})+\frac{3297}{64}\). From this, you can find that the maximum is 3297/64 ft, and it occurs after 15/32 sec.

Answer 6

Sorry for taking so long; the numbers were not very nice :D

Answer 7

Do you know how to complete a square?

Answer 8

It's okay! And, no i do not.

Answer 9

First take -16 as a common factor \(-16(t^2-\frac{15}{16}t)+48 \) (*). Now, consider the part \(t^2-\frac{15}{16}t\). Add to it the square of its half, that's \((\frac{15}{32})^2\). That makes it \(t^2-\frac{15}{16}t+(\frac{32}{16})^2=(t-\frac{15}{16})^2\). Put that back in our equation (*), but we have to subtract the amount we added. So, it becomes \(-16(t-\frac{15}{16})^2+48-(\frac{32}{16})^2\).

Answer 10

These numbers are quite large and have a lot of fractions. It's usually much simpler than this.