## Mathhelp346 3 years ago PLEASEE help with attached: How do I complete the table shown on the second picture?

1. Mathhelp346

2. Mathhelp346

Activity 1 is the first picture

3. anonymous

what is the velocity of projection???

4. Mathhelp346

5. Mathhelp346

Does that help?

6. anonymous

yep

7. anonymous

do u know the equations of motion??

8. Mathhelp346

What do you mean?

9. anonymous

okay leave it

10. Mathhelp346

Ok

11. anonymous

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

12. anonymous

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. anonymous

no do u know the formula V = U + gt

15. Mathhelp346

No

16. anonymous

u r of which standard??

17. Mathhelp346

?

18. anonymous

class of study?

19. Mathhelp346

20. anonymous

do u know about displacement,velocity,acceleration etc

21. Mathhelp346

Yes

22. anonymous

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. anonymous

now ur first question maximum height of the ball

25. anonymous

at maximum heicht, final velocity is zero

26. Mathhelp346

Ok

27. anonymous

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

28. anonymous

here acceleration is nothing but acceleration due to gravity

29. anonymous

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

30. anonymous

so finally t =U/g

31. Mathhelp346

What do you mean acceleration is nothing?

32. anonymous

acceleration is equal to acceleration due to gravity in this case

33. Mathhelp346

Ok

34. anonymous

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

35. Mathhelp346

Ok

36. anonymous

so u can do this on remaining planets

37. Mathhelp346

So Mercury would be 2 seconds right

38. anonymous

2.02

39. anonymous

next maximum height reached by the ball

40. Mathhelp346

.85 seconds

41. anonymous

for what?

42. Mathhelp346

Venus

43. anonymous

correct

44. Mathhelp346

Ok

45. anonymous

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. anonymous

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. anonymous

ya

52. Mathhelp346

Ok

53. anonymous

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. anonymous

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

57. anonymous

ya right 27.32

58. Mathhelp346

Ok

59. anonymous

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

60. Mathhelp346

Ok

61. anonymous

62. anonymous

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

65. anonymous

no

66. anonymous

wait this is some what lengthy

67. anonymous

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

68. Mathhelp346

Ok

69. anonymous

but the ball should reach the ground

70. anonymous

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. anonymous

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

72. anonymous

now it should reach the ground by traveling 3 feet

73. Mathhelp346

So 27 feet

74. anonymous

no how do u say that?

75. anonymous

u der?

76. Mathhelp346

Yes

77. anonymous

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

78. anonymous

t = 0.116 seconds

79. anonymous

80. anonymous

hope u understood

81. Mathhelp346

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

82. anonymous

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

83. anonymous

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. anonymous

ya

86. Mathhelp346

So does that mean they all travel 3 feet downward?

87. anonymous

no

88. Mathhelp346

but the ho for each of them is 3

89. anonymous

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. anonymous

same 3 feet

92. Mathhelp346

So the other ones travel 3 feet downward?

93. anonymous

|dw:1355813142620:dw|

94. Mathhelp346

Ok

95. anonymous

i think now u can understand clearly

96. anonymous

okay?

97. Mathhelp346

Ok thank you

98. anonymous

welcome

99. anonymous

:)

100. Mathhelp346

Mercury's time would be 4 seconds?