## Mathhelp346 2 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

what is the velocity of projection???

4. Mathhelp346

5. Mathhelp346

Does that help?

yep

do u know the equations of motion??

8. Mathhelp346

What do you mean?

okay leave it

10. Mathhelp346

Ok

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

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?

no do u know the formula V = U + gt

15. Mathhelp346

No

u r of which standard??

17. Mathhelp346

?

class of study?

19. Mathhelp346

do u know about displacement,velocity,acceleration etc

21. Mathhelp346

Yes

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

now ur first question maximum height of the ball

at maximum heicht, final velocity is zero

26. Mathhelp346

Ok

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

here acceleration is nothing but acceleration due to gravity

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

so finally t =U/g

31. Mathhelp346

What do you mean acceleration is nothing?

acceleration is equal to acceleration due to gravity in this case

33. Mathhelp346

Ok

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

35. Mathhelp346

Ok

so u can do this on remaining planets

37. Mathhelp346

So Mercury would be 2 seconds right

2.02

next maximum height reached by the ball

40. Mathhelp346

.85 seconds

for what?

42. Mathhelp346

Venus

correct

44. Mathhelp346

Ok

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

46. Mathhelp346

Ok

47. Mathhelp346

Wait how do you find the maximum height?

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?

ya

52. Mathhelp346

Ok

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

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

ya right 27.32

58. Mathhelp346

Ok

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

60. Mathhelp346

Ok

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

no

wait this is some what lengthy

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

68. Mathhelp346

Ok

but the ball should reach the ground

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

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

now it should reach the ground by traveling 3 feet

73. Mathhelp346

So 27 feet

no how do u say that?

u der?

76. Mathhelp346

Yes

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

t = 0.116 seconds

hope u understood

81. Mathhelp346

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

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

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?

ya

86. Mathhelp346

So does that mean they all travel 3 feet downward?

no

88. Mathhelp346

but the ho for each of them is 3

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?

same 3 feet

92. Mathhelp346

So the other ones travel 3 feet downward?

|dw:1355813142620:dw|

94. Mathhelp346

Ok

i think now u can understand clearly

okay?

97. Mathhelp346

Ok thank you

welcome