KJ4UTS one year ago Suppose a rock is thrown upwards from the top of a building 200 feet high with an initial velocity of 96 feet per second. The height (in feet) of the rock above the ground t seconds after being released is given by the formula h(t) = -16t^2 + 96t + 200 1.When will it hit the ground? 2.How high will it be after 2.1 seconds? 3.What is the maximum height?

1. anonymous

1. It will hit the ground when h(t) = 0 so you need to solve $-16t^2+96t+200=0$

2. anonymous

3. KJ4UTS

Would it be t=3-√ 43/2=-1.63 and 3+√ 43/2=7.63

4. anonymous

yes, and since time can't be negative, it hits the ground at 7.63 seconds.

5. anonymous

For 2, plug in 2.1 for t to find h(2.1).

6. KJ4UTS

331.04

7. anonymous

yes. and then for 3 there are two steps. Use $$t=-\frac{ b }{ 2a }$$ to find the time of max height. Then plug that time into the h(t) equation

8. KJ4UTS

Im a little confused as to what I plug into b and a?

9. anonymous

the same values you used in the quadratic equation a = -16, b = 96, c = 200

10. anonymous

$t=-\frac{ 96 }{ 2(-16) }$

11. KJ4UTS

3

12. anonymous

now plug 3 in for t in the original equation

13. KJ4UTS

Ok I got 344 thank you for explaining this step by step :)

14. anonymous

yeah that's it. you're welcome. :)