## anonymous one year ago s(t)=-16x^2+80t+100

1. zepdrix

$\large\rm s(t)=-16t^2+80t+100$What's up Cheyenne? :) What do you need to do with this function?

2. anonymous

This is the problem : 1. A ball is thrown upward from an initial height of 100 ft with an initial velocity of 80 ft per sec. The path of the ball is modeled by the function: After how many seconds does the projectile reach its maximum height? What is this maximum height? After how many seconds will the ball hit the ground?

3. zepdrix

The shape of the path the ball follows is parabolic. So the maximum height is going to be located at the vertex of the parabola. Do you remember how to put a parabola into vertex form? :)

4. anonymous

no I do not

5. zepdrix

$\large\rm -16t^2+80t\qquad\qquad +100$Ignore the +100 for now. We want to try and complete the square on the t's. We'll start by factoring -16 out of each term to get the coefficient off of the squared term.$\large\rm -16(\color{orangered}{t^2-5t})\qquad\qquad +100$And now we want to try and complete the square on this orange part.

6. zepdrix

In case there in any confusion on what I just did: I divide -16 out of 16t^2, leaving t^2 in the first part of the bracket. And I divided -16 out of 80 leaving a 5 on the t. And the -16 is multiplying the outside of the brackets.

7. zepdrix

So uhhh to complete the square :p let's see...

8. anonymous

okay i get the dividing out the -16 part

9. zepdrix

Let's split the t's into two equal groups, not the squares but the other ones. $\large\rm t^2-5t\quad=\quad t^2-2.5t-2.5t$We can draw a square to show this relationship. The side length will be t-2.5.|dw:1441252166003:dw|

10. zepdrix

|dw:1441252302630:dw|If we fill in the area of the pieces, we can see that to complete the square we need the bottom right corner piece. Which is -2.5 * -2.5 in area.

11. anonymous

6.25 ?

12. zepdrix

If the squares are confusing, we can just stick to the formulas :) The formula tells you to take half of your b coefficient, and square it. So our b value is -5, half of that is -2.5, squaring gives us 6.25. Good good good, that's the value that completes the square for us.

13. zepdrix

$\large\rm -16(\color{orangered}{t^2-5t+6.25})\qquad\qquad +100$But we can't just add 6.25 willy nilly like this. We'll keep the equation balanced by also subtracting 6.25.$\large\rm -16(\color{orangered}{t^2-5t+6.25}-6.25)\qquad\qquad +100$

14. zepdrix

But the -6.25 isn't part of our perfect square, we want to take it outside of the brackets. See how there is a -16 multiplying that entire bracket? We have to multiply the value by -16 to take it outside.

15. zepdrix

$\large\rm =-16(\color{orangered}{t^2-5t+6.25})-16(-6.25)+100$

16. zepdrix

And our perfect square condenses down to: $$\large\rm \left(t+\frac{b}{2}\right)^2$$ So we end up with:$\large\rm =-16(t+2.5)^2-16(-6.25)+100$

17. zepdrix

Then you need to combine the numbers on the outside to find the y-coordinate of your vertex. $$\large\rm f(x)=(x-h)^2+k$$ Where h is the x-coordinate of your vertex and k is the y-coordinate of your vertex. The y-coordinate of your vertex corresponds to the height at that moment.

18. zepdrix

What do you think Chey? :d too confusing?

19. anonymous

Im kinda understanding it my teacher didnt really teach us the vertex from she just thought we knew it

20. zepdrix

So at time t=2.5, the ball will reach it's highest height of $$\large\rm -16(-6.25)+100$$ feet. So how high up is that? :) Come on Chey! You can do the last calculation! :D

21. anonymous

200 ft

22. zepdrix

yay good job \c:/ That takes care of the first part.

23. zepdrix

Oh actually that's the second question.. I kinda spilled the beans on the first part :P it's the x-coordinate of the vertex, 2.5 seconds.

24. zepdrix

The third question asks, After how many seconds will the ball hit the ground? So when the ball is on the ground... that corresponds to a height of zero... since it's in the ground. And remember, the function, s(t), represents height. $\large\rm s(t)=-16t^2+80t+100$So we let the height be zero.$\large\rm 0=-16t^2+80t+100$And we solve for t to figure out at what time this actually happens.

25. zepdrix

Do you understand how to solve for t? Try factoring, if that's too confusing, you can always turn to your quadratic function.$\large\rm t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

26. zepdrix

27. anonymous

so you would take out a -4 . so then it would be $-4(4t^2-20t-25)$

28. zepdrix

Mmm ya you could do that :)$\large\rm 0=-4(4t^2-20t-25)$And then divide both sides by -4,$\large\rm 0=4t^2-20t-25$That certainly simplifies it a tad bit.

29. anonymous

so then do i factor the $0=(4t^2-20t-25)$

30. zepdrix

Oh it appears it won't factor :\ So let's save ourselves the trouble of even trying. Just plug the numbers into your quadratic formula:$\large\rm t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$and simplify! :)

31. anonymous

so would the 20 be positive or negative for the -b

32. zepdrix

$\large\rm t=\frac{-(b)\pm\sqrt{(b)^2-4ac}}{2a}$You're plugging a -20 into the b,$\large\rm t=\frac{-(-20)\pm\sqrt{(-20)^2-4ac}}{2a}$So I guess that first one is going to be positive, ya?

33. anonymous

okay so then what ?

34. zepdrix

Le sigh.. :) Plug the other values in silly!$\large\rm t=\frac{-(-20)\pm\sqrt{(-20)^2-4(4)(-25)}}{2(4)}$And simplify! :D

35. anonymous

36. anonymous

so the $20^2 = 400 and the -4 *4*-25 equals 400$

37. anonymous

Do you mind helping with a different problem ?

38. zepdrix

Ya so you get uhhhhh$\large\rm t=\frac{-(-20)\pm\sqrt{400-(-100)}}{2(4)}$$\large\rm t=\frac{20\pm\sqrt{400+400}}{8}$$\large\rm t\approx\frac{20\pm28.28}{8}$Where the positive root is the only one that gives us a positive t value. $\large\rm t\approx\frac{20+28.28}{8}$Which gives you like 6ish. something like this ya? :o

39. zepdrix

A different problem? >.< Grr I have so much homework to dooooo :O Close down this thread, its getting too long and messy. Open a new one with your new question. You can type @zepdrix to page me, I'll come take a look if I have time :DDD