Mathhelp346
PLEASEE help with attached:
How do I complete the table shown on the second picture?
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Mathhelp346
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Mathhelp346
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Activity 1 is the first picture
bhaskarbabu
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what is the velocity of projection???
Mathhelp346
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Mathhelp346
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Does that help?
bhaskarbabu
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yep
bhaskarbabu
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do u know the equations of motion??
Mathhelp346
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What do you mean?
bhaskarbabu
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okay leave it
Mathhelp346
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Ok
bhaskarbabu
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time taken to reach the maximum height t = u/g
bhaskarbabu
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on earth surface , t = 24/32
=3/4 second
Mathhelp346
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So it's like \[v=\frac{ d }{ t }\]Where d is distance and t is time?
bhaskarbabu
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no do u know the formula
V = U + gt
Mathhelp346
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No
bhaskarbabu
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u r of which standard??
Mathhelp346
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?
bhaskarbabu
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class of study?
Mathhelp346
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8th grade?
bhaskarbabu
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do u know about displacement,velocity,acceleration etc
Mathhelp346
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Yes
bhaskarbabu
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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
Mathhelp346
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Ok
bhaskarbabu
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now ur first question
maximum height of the ball
bhaskarbabu
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at maximum heicht,
final velocity is zero
Mathhelp346
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Ok
bhaskarbabu
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so,
0=U+at
t = -U/a
bhaskarbabu
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here acceleration is nothing but acceleration due to gravity
bhaskarbabu
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when a body is projected vertically upwards acceleation due to gravity is taken negetive
bhaskarbabu
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so finally t =U/g
Mathhelp346
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What do you mean acceleration is nothing?
bhaskarbabu
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acceleration is equal to acceleration due to gravity in this case
Mathhelp346
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Ok
bhaskarbabu
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so finally time required to reach maximum height t = U/g
= 24/32
=3/4 second
Mathhelp346
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Ok
bhaskarbabu
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so u can do this on remaining planets
Mathhelp346
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So Mercury would be 2 seconds right
bhaskarbabu
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2.02
bhaskarbabu
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next maximum height reached by the ball
Mathhelp346
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.85 seconds
bhaskarbabu
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for what?
Mathhelp346
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Venus
bhaskarbabu
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correct
Mathhelp346
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Ok
bhaskarbabu
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do the remaining later
lt's go for the next question
Mathhelp346
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Ok
Mathhelp346
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Wait how do you find the maximum height?
bhaskarbabu
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for maximum height reached substitute t = U/g in h(t)
Mathhelp346
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You said it was t = -U/a?
Mathhelp346
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So I substitute the time required to reach maximum height into the quadratic functions?
bhaskarbabu
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ya
Mathhelp346
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Ok
bhaskarbabu
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u will get h(t) = U^2/2g + ho
Mathhelp346
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Ok
Mathhelp346
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So the answer to the first one would be 27 ft right
bhaskarbabu
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now on earth surface,
h = (24)^2/2(32) +3
= 12
bhaskarbabu
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ya right 27.32
Mathhelp346
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Ok
bhaskarbabu
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now u can do the remaining
let's move to the next question
Mathhelp346
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Ok
bhaskarbabu
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time required to return to planet's surface
bhaskarbabu
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if the ball is projected from the ground
time taken to reach the maximum height = time taken to reach the ground from maximum height
Mathhelp346
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Ok
Mathhelp346
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So it's the same answer?
bhaskarbabu
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no
bhaskarbabu
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wait this is some what lengthy
bhaskarbabu
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now, time taken by the ball to reach the hand of the person is,
U/g + U/g = 2U/g
Mathhelp346
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Ok
bhaskarbabu
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but the ball should reach the ground
bhaskarbabu
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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
bhaskarbabu
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so it reaches the hands of the person with a velocity of 24 feet per second
bhaskarbabu
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now it should reach the ground by traveling 3 feet
Mathhelp346
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So 27 feet
bhaskarbabu
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no how do u say that?
bhaskarbabu
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u der?
Mathhelp346
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Yes
bhaskarbabu
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now h(t) = (1/2)g(t^2) + Ut
3 = 16t^2 + 24t
bhaskarbabu
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t = 0.116 seconds
bhaskarbabu
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now add this with 2U/g
bhaskarbabu
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hope u understood
Mathhelp346
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There's a negative in front of (1/2)g
And how did you get 3?
bhaskarbabu
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that is given because there the ball is projected vertically upward
here the ball is coming down
bhaskarbabu
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it should travel 3 feet downward to reach the ground
Mathhelp346
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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?
bhaskarbabu
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ya
Mathhelp346
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So does that mean they all travel 3 feet downward?
bhaskarbabu
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no
Mathhelp346
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but the ho for each of them is 3
bhaskarbabu
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the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground
Mathhelp346
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So how long do the other ones travel?
bhaskarbabu
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same 3 feet
Mathhelp346
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So the other ones travel 3 feet downward?
bhaskarbabu
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|dw:1355813142620:dw|
Mathhelp346
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Ok
bhaskarbabu
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i think now u can understand clearly
bhaskarbabu
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okay?
Mathhelp346
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Ok thank you
bhaskarbabu
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welcome
bhaskarbabu
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:)
Mathhelp346
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Mercury's time would be 4 seconds?