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PLEASEE help with attached: How do I complete the table shown on the second picture?

Mathematics
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Activity 1 is the first picture
what is the velocity of projection???

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Other answers:

1 Attachment
Does that help?
yep
do u know the equations of motion??
What do you mean?
okay leave it
Ok
time taken to reach the maximum height t = u/g
on earth surface , t = 24/32 =3/4 second
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
No
u r of which standard??
?
class of study?
8th grade?
do u know about displacement,velocity,acceleration etc
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
Ok
now ur first question maximum height of the ball
at maximum heicht, final velocity is zero
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
What do you mean acceleration is nothing?
acceleration is equal to acceleration due to gravity in this case
Ok
so finally time required to reach maximum height t = U/g = 24/32 =3/4 second
Ok
so u can do this on remaining planets
So Mercury would be 2 seconds right
2.02
next maximum height reached by the ball
.85 seconds
for what?
Venus
correct
Ok
do the remaining later lt's go for the next question
Ok
Wait how do you find the maximum height?
for maximum height reached substitute t = U/g in h(t)
You said it was t = -U/a?
So I substitute the time required to reach maximum height into the quadratic functions?
ya
Ok
u will get h(t) = U^2/2g + ho
Ok
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
Ok
now u can do the remaining let's move to the next question
Ok
time required to return to planet's surface
if the ball is projected from the ground time taken to reach the maximum height = time taken to reach the ground from maximum height
Ok
So it's the same answer?
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
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
So 27 feet
no how do u say that?
u der?
Yes
now h(t) = (1/2)g(t^2) + Ut 3 = 16t^2 + 24t
t = 0.116 seconds
now add this with 2U/g
hope u understood
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
\[h(t)=-\frac{ 1 }{ 2 }g t ^{2}+v _{0}t +h _{0}\] So you got the 3 from Earth's quadratic function right?
ya
So does that mean they all travel 3 feet downward?
no
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
So how long do the other ones travel?
same 3 feet
So the other ones travel 3 feet downward?
|dw:1355813142620:dw|
Ok
i think now u can understand clearly
okay?
Ok thank you
welcome
:)
Mercury's time would be 4 seconds?

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