## A community for students. Sign up today

Here's the question you clicked on:

## ChrisV 3 years ago s(t)=-4.9+vt+s To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 5.6 seconds after the stone is dropped?

• This Question is Closed
1. ChrisV

s(t)=-4.9t+vt+s

2. Awstinf

hmm are you aloud to use derivatives?

3. ChrisV

missed a t >.<

4. Awstinf

-4.9t^2 I think you mean :)

5. ChrisV

yea

6. ChrisV

-4.9t^2 +vt+s

7. ChrisV

and yes i can use derivatives

8. ChrisV

im just not sure how to use the information given

9. ChrisV

im so confused right now

10. ChrisV

this rate of change thing is going to do me in

11. Awstinf

hmm so vo and so are both unknown?

12. ChrisV

yes but i am assuming s0 would = 0 at 5.6 seconds

13. ChrisV

since thats impact

14. Awstinf

yeah thats right the vo is just the pain in the a part of the problem

15. Awstinf

wait wait wait, it seems to be lacking the classic physics wording of it was dropped with an unknown intial velocity, it being dropped and not thrown I would maybe go as far to say the the intial velocity is 0

16. Awstinf

that seems like it might be too easy though

17. Awstinf

we dont drop things with an intial velocity though lol, thats against the definition of dropping lol

18. ChrisV

I am so lost on any problems dealing with rate of change

19. ChrisV

I may have to drop this course if I do not pick these up soon :(

20. Awstinf

:(

21. Awstinf

4.9(5.6^2) should be your answer assuming that dropping means letting go from rest

22. ChrisV

I am having a hard time telling where the information they give me plugs in

23. ChrisV

Say they give me a distance of 108 meters, I assume thats s

24. ChrisV

if they give me a velocity that v0t

25. ChrisV

and a measurement of time = t

26. ChrisV

so if they give me t and vot

27. ChrisV

i find the derivative

28. ChrisV

and solve for the value of t

29. ChrisV

but when they give me t and s im lost lol

30. Awstinf

s(t)= -4.9t^2+v0t+108?

31. ChrisV

and when they only give me t i am more than lost

32. ChrisV

no that more like my next problem, but if i were given that question

33. Awstinf

Here ill read type you a problem from my book and try to help clarify things

34. ChrisV

i do not know what to do

35. Awstinf

read-type*

36. Awstinf

Hokay so, A person standing at the edge of a cliff throws a rock directly upward. it is observed that 2 seconds later the rock is at its maximum height and that 5 seconds after that, it hits the ground at the base of the cliff?

37. Awstinf

questions like that?

38. ChrisV

i have no idea how to solve that

39. Awstinf

But they are questions like that correct?

40. Awstinf

I'll show you how to appoarch it if it is

41. Awstinf

(this is straight from my change of rate section)

42. ChrisV

hmmmm on my original question

43. Awstinf

yeah?

44. ChrisV

s(t)=-4.9+vt+s v(t)=s'(t)

45. ChrisV

oops 4.9t^2

46. ChrisV

so -9.8t+v would be the derivative

47. Awstinf

yeah

48. ChrisV

-9.8(5.6)+v=0 v= 54.88

49. ChrisV

would that be the velocity?

50. Awstinf

thats your impact velocity

51. ChrisV

ok so when s=0 Velocity = 54.88?

52. Awstinf

assuming that v0=0 yes

53. ChrisV

now i confused myself lol

54. Awstinf

hahaha remember the derivative of the distance equation is the velocity at time equation

55. ChrisV

-4.9t^2 +vt+s

56. ChrisV

well if s=distance

57. ChrisV

at 5.6 secs s=0 since thats impact right?

58. Awstinf

yes

59. Awstinf

wait nno no s0 is not o

60. Awstinf

0*

61. Awstinf

here label the equation like this it makes it more clear. So 4.9t^2+v0t+s0= s

62. Awstinf

so at time t=5.6s s would be 0 not s0

63. Awstinf

other wise you would get a negative height, it really doesnt matter which way you do it but you should do it right when you are first doing it.

64. ChrisV

hehe

65. ChrisV

well i am trying to understand

66. Awstinf

I'm sorry I really havent been too much help but that question is horribly worded. If I were in your shoes I would assume that the initial velocity is 0 as progress from there. As long as you understand the relationships between the equations, it will make solving these very simple. Drawing whats going on helps

67. ChrisV

-4.9t^2+vt+s 9.8t+v is the derivative correct?

68. Awstinf

yes sir

69. ChrisV

with that i can get the velocity at impact

70. ChrisV

-9.8(5.6)+v=0 v= 54.88

71. Awstinf

yes but you need to know what v initial is

72. ChrisV

well wouldnt that be as simple as making t 0?

73. Awstinf

I dont think you can do that because you are stating that v(5.6)=0

74. Awstinf

Its hard to explain

75. ChrisV

yea i see lol

76. Awstinf

Try reposting this in the physics section, they are really good at explain this kind of stuff

77. Awstinf
78. ChrisV

will do thanks

79. Awstinf

I was right :)

80. Awstinf

the initial velocity is 0

81. ChrisV

:)

82. ChrisV

:)

83. Awstinf

Sorry I was trying to be ambiguous I didnt want to give you improper information

84. Awstinf

I am in your same level class lol

85. Awstinf

I just do about 6 hours of hw a day

86. ChrisV

i took too many courses

87. ChrisV

falling behind

88. Awstinf

yeah it blows I have no social life atm just study study

89. ChrisV

com 101 is more time consuming than i thought

90. Awstinf

:( yeah thats the draw back of underestimating courses. I'm taking physics, chem, physics lab, calc and I really thought this semester would be a tad less stressful

91. ChrisV

chem, com 101, calc and us history

92. ChrisV

history and chem is ok, but com 101 is taking way to much of my time

93. Awstinf

us history=gun to my head i finished all of my gen ed first so I could figure out what I wanna do

94. ChrisV

taking way to much time writing and preparing for speeches

95. Awstinf

Thats why I stick to courses that strictly deal with numbers

96. Awstinf

lol

97. ChrisV

lol its a req for my degree or i would never touch it

98. Awstinf

:( I just wanna take comparitive virology already, all of these courses building up to it are just BS

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy