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## Mathhelp346 Group Title PLEASEE help with attached: How do I complete the table shown on the second picture? one year ago one year ago

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1. Mathhelp346

2. Mathhelp346

Activity 1 is the first picture

3. bhaskarbabu

what is the velocity of projection???

4. Mathhelp346

5. Mathhelp346

Does that help?

6. bhaskarbabu

yep

7. bhaskarbabu

do u know the equations of motion??

8. Mathhelp346

What do you mean?

9. bhaskarbabu

okay leave it

10. Mathhelp346

Ok

11. bhaskarbabu

time taken to reach the maximum height t = u/g

12. bhaskarbabu

on earth surface , t = 24/32 =3/4 second

13. Mathhelp346

So it's like $v=\frac{ d }{ t }$Where d is distance and t is time?

14. bhaskarbabu

no do u know the formula V = U + gt

15. Mathhelp346

No

16. bhaskarbabu

u r of which standard??

17. Mathhelp346

?

18. bhaskarbabu

class of study?

19. Mathhelp346

8th grade?

20. bhaskarbabu

do u know about displacement,velocity,acceleration etc

21. Mathhelp346

Yes

22. bhaskarbabu

okay so now u learn this formula V= U + at V= final velocity of the body U= initial velocity of the body a = acceleration t = time

23. Mathhelp346

Ok

24. bhaskarbabu

now ur first question maximum height of the ball

25. bhaskarbabu

at maximum heicht, final velocity is zero

26. Mathhelp346

Ok

27. bhaskarbabu

so, 0=U+at t = -U/a

28. bhaskarbabu

here acceleration is nothing but acceleration due to gravity

29. bhaskarbabu

when a body is projected vertically upwards acceleation due to gravity is taken negetive

30. bhaskarbabu

so finally t =U/g

31. Mathhelp346

What do you mean acceleration is nothing?

32. bhaskarbabu

acceleration is equal to acceleration due to gravity in this case

33. Mathhelp346

Ok

34. bhaskarbabu

so finally time required to reach maximum height t = U/g = 24/32 =3/4 second

35. Mathhelp346

Ok

36. bhaskarbabu

so u can do this on remaining planets

37. Mathhelp346

So Mercury would be 2 seconds right

38. bhaskarbabu

2.02

39. bhaskarbabu

next maximum height reached by the ball

40. Mathhelp346

.85 seconds

41. bhaskarbabu

for what?

42. Mathhelp346

Venus

43. bhaskarbabu

correct

44. Mathhelp346

Ok

45. bhaskarbabu

do the remaining later lt's go for the next question

46. Mathhelp346

Ok

47. Mathhelp346

Wait how do you find the maximum height?

48. bhaskarbabu

for maximum height reached substitute t = U/g in h(t)

49. Mathhelp346

You said it was t = -U/a?

50. Mathhelp346

So I substitute the time required to reach maximum height into the quadratic functions?

51. bhaskarbabu

ya

52. Mathhelp346

Ok

53. bhaskarbabu

u will get h(t) = U^2/2g + ho

54. Mathhelp346

Ok

55. Mathhelp346

So the answer to the first one would be 27 ft right

56. bhaskarbabu

now on earth surface, h = (24)^2/2(32) +3 = 12

57. bhaskarbabu

ya right 27.32

58. Mathhelp346

Ok

59. bhaskarbabu

now u can do the remaining let's move to the next question

60. Mathhelp346

Ok

61. bhaskarbabu

time required to return to planet's surface

62. bhaskarbabu

if the ball is projected from the ground time taken to reach the maximum height = time taken to reach the ground from maximum height

63. Mathhelp346

Ok

64. Mathhelp346

So it's the same answer?

65. bhaskarbabu

no

66. bhaskarbabu

wait this is some what lengthy

67. bhaskarbabu

now, time taken by the ball to reach the hand of the person is, U/g + U/g = 2U/g

68. Mathhelp346

Ok

69. bhaskarbabu

but the ball should reach the ground

70. bhaskarbabu

there one more property that, if a ball is projected vertically up with certain velocity ,the ball acquire the same velocity when it reaches the original position

71. bhaskarbabu

so it reaches the hands of the person with a velocity of 24 feet per second

72. bhaskarbabu

now it should reach the ground by traveling 3 feet

73. Mathhelp346

So 27 feet

74. bhaskarbabu

no how do u say that?

75. bhaskarbabu

u der?

76. Mathhelp346

Yes

77. bhaskarbabu

now h(t) = (1/2)g(t^2) + Ut 3 = 16t^2 + 24t

78. bhaskarbabu

t = 0.116 seconds

79. bhaskarbabu

now add this with 2U/g

80. bhaskarbabu

hope u understood

81. Mathhelp346

There's a negative in front of (1/2)g And how did you get 3?

82. bhaskarbabu

that is given because there the ball is projected vertically upward here the ball is coming down

83. bhaskarbabu

it should travel 3 feet downward to reach the ground

84. Mathhelp346

$h(t)=-\frac{ 1 }{ 2 }g t ^{2}+v _{0}t +h _{0}$ So you got the 3 from Earth's quadratic function right?

85. bhaskarbabu

ya

86. Mathhelp346

So does that mean they all travel 3 feet downward?

87. bhaskarbabu

no

88. Mathhelp346

but the ho for each of them is 3

89. bhaskarbabu

the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground

90. Mathhelp346

So how long do the other ones travel?

91. bhaskarbabu

same 3 feet

92. Mathhelp346

So the other ones travel 3 feet downward?

93. bhaskarbabu

|dw:1355813142620:dw|

94. Mathhelp346

Ok

95. bhaskarbabu

i think now u can understand clearly

96. bhaskarbabu

okay?

97. Mathhelp346

Ok thank you

98. bhaskarbabu

welcome

99. bhaskarbabu

:)

100. Mathhelp346

Mercury's time would be 4 seconds?