anonymous one year ago 4) What is the height of the apple when your friend catches it? (**HINT: The height would be the y-coordinate of the highest point.) Double click on the space below to show your work. h(t) = -16t^2 + 38.4t + 0.96 h= height t= time (seconds) I will fan and medal for work plus answers!

1. anonymous

How do I find the height? How do I switch the equation to find the height instead of the time? I can solve it just not sure how to set it up.

2. anonymous

there is more to this question right?

3. anonymous

like more words

4. anonymous

Make t=0. :)

5. anonymous

The information is enough. :)

6. anonymous

naw dude its asking for the vertex

7. anonymous

Would I just substitute the time that we just figured out into the equation? The 2.4 seconds?

8. anonymous

|dw:1444445933109:dw|

9. anonymous

No you don't have to. Just make t=0. :) $$h(t) = -16(0)2 + 38.4(0) + 0.96$$

10. anonymous

why would you do that?? is the friend .96 ft off the ground

11. anonymous

The height we're finding is the same thing as the vertex. =_= Notice that the vertex in this case is the maximum because $$a$$ is a negative.

12. anonymous

no not for t=0

13. anonymous

t=o is your initial starting point not the vertex

14. anonymous

15. anonymous

If (0,y) <--- this is the vertex. Please stop misguiding the one who asked.

16. anonymous

NO IT ISNT...

17. anonymous

If you solve the equation mathway gave me, I got 0.96.

18. anonymous

b/2a is the x coordinate and f(b/2a) is your Y

19. anonymous

"(**HINT: The height would be the y-coordinate of the highest point.)" Which means that you need to find the y-coordinate.

20. anonymous

Yes, it is $$\huge 0.96$$. :)

21. anonymous

MAX HIGHT WHICH IS THE Y OF YOUR VERTEX

22. anonymous

t=0 is not a vertex

23. anonymous

that is 0 time e=meaning the apple hasnt been thrown yet...

24. anonymous

It is not a vertex because it is PART OF THE VERTEX. It is the x-intercept.

25. anonymous

how can your friend catch something that hasnt been thrown?

26. anonymous

27. anonymous

no x intercept is finding the root dude...

28. anonymous

you're really off

29. anonymous

@pooja195

30. anonymous

The question I posted before this was asking if a person threw an apple how many seconds would it take to reach the other person 3 stories up? We solved and got 2.4 seconds.

31. anonymous

yes

32. anonymous

Show me how you will find the height then. You keep on complaining, but you're not showing any work.

33. anonymous

Of course it is correct. I was the one who helped her. :P

34. Nnesha

(**HINT: The height would be the y-coordinate of the highest point.)

35. anonymous

but now plug that number into the function

36. anonymous

that is the hight

37. anonymous

not t=0...

38. anonymous

@mathway im not calling you dumb people make mistakes and this is one of yours

39. anonymous

You would still get the same answer either way, 0.96.

40. anonymous

yes buy he cant say thats what it is you have to explain it in the context of the problem

41. anonymous

since its a root ofcourse it would cancle out the other 2 but in a 4th power polynomial that would fail miserably

42. anonymous

so the fundamental way to explain the answer is V = {(b/2a),f(b/2a)} and you look for Y

43. anonymous

Okay guys last question! Using the same equation - h(t) = -16t^2 + 38.4t + 0.96 With h = height and t = time in seconds 5) Suppose your friend does not catch the apple and it falls back to the ground. How long did it take the apple to hit the ground? (**HINT: When the apple hits the ground is when the graph hits the x-axis.)

44. anonymous

find the root

45. anonymous

|dw:1444446732488:dw|

46. Nnesha

calm plz use the hint highest point = max so we need y -coordinate of the max point (max=vertex)

47. Nnesha

lol sorry for late reply >.< internet iz very slow -.-

48. anonymous

lol its ok

49. anonymous

ok hold up go back what did you get for your sec for the apple to get to him??????!!!!

50. anonymous

the kid on the 3rd floor????

51. Nnesha

x coordinate of the vertex can be any number (0,y) = y-intercept :=)

52. anonymous

$\frac{ -38.4 \pm \sqrt{38.4^{2}-4(-16)(0.96)} }{ 2(-16) }$

53. anonymous

yea lol but only if it is centered on the orgin would that be a vertex

54. anonymous

yea @tbull1221 that works

55. anonymous

what does that give you?

56. anonymous

$\frac{ -38.4 \pm \sqrt{1536} }{ -32 }$

57. anonymous

Then you get: $\frac{ -38.4 \pm 16\sqrt{6} }{ -32 }$

58. anonymous

"x coordinate of the vertex can be any number (0,y) = y-intercept :=)" @Nnesha Thank you for backing up! At least I know that I'm not dumb and I don't necessarily have to say irrelevant things when I made a mistake. :)

59. anonymous

btw dude the other answer for the seconds it takes was wrong its 4.8sec

60. anonymous
61. anonymous

im in 3rd year calc and 2 year stat i know how to find the vertex of a parabola...

62. anonymous

Can someone please just give me an answer for #5? My computer is about to die. I will fan and medal if i haven't already.

63. anonymous

I highly suggest that you don't call me "dude." We are not close.

64. anonymous

And is the seconds answer 2.4 or 4.8?

65. anonymous

sorry im a fairly welcoming person i dont show nor mean it in any form of disrespect

66. anonymous

the one that said "the amount of seconds for the apple to reach the friend" is 4.8 sec

67. Nnesha

people make mistakes i do all the time everyday more than 100 times it's okay we can correct it nicely :=) so please :=)

68. anonymous

because -b/2a =4.8

69. anonymous

70. anonymous

then plug that into your equation and you will get the amount of feet it took for the friend to get the apple

71. anonymous

then for this what you did was fine you just plug the numbers into the quadratic equation and you will get the time it took to hit the ground

72. anonymous

he got offline...

73. anonymous

do you understand why you were wrong @mathway ??

74. anonymous

not trying to disrespect you. at this point im trying to help you

75. anonymous

@tbull1221 cool?

76. anonymous

go up like 5 comments and i did a step-by-step of the process

77. anonymous

So the seconds is 4.8? And What is the answer to #5?

78. anonymous

5 is the ft??

79. anonymous

5) Suppose your friend does not catch the apple and it falls back to the ground. How long did it take the apple to hit the ground? (**HINT: When the apple hits the ground is when the graph hits the x-axis.)

80. anonymous

but both of the ones before that were wrong because the sec were wrong so the FT to make to him wouldnt be .96

81. anonymous

so with the answer to -b/2a plug that number into the equation given and you will get the distance of the throw to the vertex (the friend)

82. anonymous

because mathway said it was 2.4 sec but it wasnt it was 4.8

83. anonymous

so then plug in 4.8 into the equation to get the answer to the Max height

84. anonymous

then what you did with the quadratic equation would give you #5

85. anonymous

Cool? @tbull1221

86. anonymous

@tbull1221 are you still there??

87. anonymous

aww tbull got off :(

88. Jhannybean

\begin{align} h(t) = -16t^2 + 38.4t + 0.96 &\implies h(t) = -16t^2+\frac{192}{5}t+\frac{24}{25} \\&\implies h(t) =t^2 -\frac{12}{5}t -\frac{3}{50} =0 \\&\implies h(t) = \left(t^2-\frac{12}{5}t \right) =\frac{3}{50} \\&\implies h(t) = \left(t^2 -\frac{12}{5}t +\frac{36}{25} \right)=\frac{3}{50}+\frac{36}{25} \\ &\implies h(t) = \left(t-\frac{6}{5}\right)^2=\frac{3}{2} \\&\implies h(t) : ~t=\frac{6}{5}\pm\sqrt{\frac{3}{2}} \end{align}

89. anonymous

yea she left:(

90. Jhannybean

time is always positive in this case, therefore $$t= \frac{6}{5} +\sqrt{\frac{3}{2}}$$