## 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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Activity 1 is the first picture

what is the velocity of projection???

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Does that help?

yep

do u know the equations of motion??

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What do you mean?

okay leave it

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Ok

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

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

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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

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No

u r of which standard??

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?

class of study?

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do u know about displacement,velocity,acceleration etc

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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

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Ok

now ur first question maximum height of the ball

at maximum heicht, final velocity is zero

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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

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What do you mean acceleration is nothing?

acceleration is equal to acceleration due to gravity in this case

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Ok

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

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Ok

so u can do this on remaining planets

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So Mercury would be 2 seconds right

2.02

next maximum height reached by the ball

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.85 seconds

for what?

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Venus

correct

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Ok

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

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Ok

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Wait how do you find the maximum height?

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

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You said it was t = -U/a?

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

ya

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Ok

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

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Ok

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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

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Ok

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

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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

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Ok

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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

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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

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So 27 feet

no how do u say that?

u der?

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Yes

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

t = 0.116 seconds

hope u understood

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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

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

ya

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So does that mean they all travel 3 feet downward?

no

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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

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So how long do the other ones travel?

same 3 feet

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So the other ones travel 3 feet downward?

|dw:1355813142620:dw|

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Ok

i think now u can understand clearly

okay?

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Ok thank you

welcome