PLEASEE help with attached:
How do I complete the table shown on the second picture?
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OpenStudy (mathhelp346):
OpenStudy (mathhelp346):
Activity 1 is the first picture
OpenStudy (anonymous):
what is the velocity of projection???
OpenStudy (mathhelp346):
OpenStudy (mathhelp346):
Does that help?
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OpenStudy (anonymous):
yep
OpenStudy (anonymous):
do u know the equations of motion??
OpenStudy (mathhelp346):
What do you mean?
OpenStudy (anonymous):
okay leave it
OpenStudy (mathhelp346):
Ok
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OpenStudy (anonymous):
time taken to reach the maximum height t = u/g
OpenStudy (anonymous):
on earth surface , t = 24/32
=3/4 second
OpenStudy (mathhelp346):
So it's like \[v=\frac{ d }{ t }\]Where d is distance and t is time?
OpenStudy (anonymous):
no do u know the formula
V = U + gt
OpenStudy (mathhelp346):
No
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OpenStudy (anonymous):
u r of which standard??
OpenStudy (mathhelp346):
?
OpenStudy (anonymous):
class of study?
OpenStudy (mathhelp346):
8th grade?
OpenStudy (anonymous):
do u know about displacement,velocity,acceleration etc
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OpenStudy (mathhelp346):
Yes
OpenStudy (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
OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
now ur first question
maximum height of the ball
OpenStudy (anonymous):
at maximum heicht,
final velocity is zero
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OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
so,
0=U+at
t = -U/a
OpenStudy (anonymous):
here acceleration is nothing but acceleration due to gravity
OpenStudy (anonymous):
when a body is projected vertically upwards acceleation due to gravity is taken negetive
OpenStudy (anonymous):
so finally t =U/g
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OpenStudy (mathhelp346):
What do you mean acceleration is nothing?
OpenStudy (anonymous):
acceleration is equal to acceleration due to gravity in this case
OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
so finally time required to reach maximum height t = U/g
= 24/32
=3/4 second
OpenStudy (mathhelp346):
Ok
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OpenStudy (anonymous):
so u can do this on remaining planets
OpenStudy (mathhelp346):
So Mercury would be 2 seconds right
OpenStudy (anonymous):
2.02
OpenStudy (anonymous):
next maximum height reached by the ball
OpenStudy (mathhelp346):
.85 seconds
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OpenStudy (anonymous):
for what?
OpenStudy (mathhelp346):
Venus
OpenStudy (anonymous):
correct
OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
do the remaining later
lt's go for the next question
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OpenStudy (mathhelp346):
Ok
OpenStudy (mathhelp346):
Wait how do you find the maximum height?
OpenStudy (anonymous):
for maximum height reached substitute t = U/g in h(t)
OpenStudy (mathhelp346):
You said it was t = -U/a?
OpenStudy (mathhelp346):
So I substitute the time required to reach maximum height into the quadratic functions?
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OpenStudy (anonymous):
ya
OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
u will get h(t) = U^2/2g + ho
OpenStudy (mathhelp346):
Ok
OpenStudy (mathhelp346):
So the answer to the first one would be 27 ft right
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OpenStudy (anonymous):
now on earth surface,
h = (24)^2/2(32) +3
= 12
OpenStudy (anonymous):
ya right 27.32
OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
now u can do the remaining
let's move to the next question
OpenStudy (mathhelp346):
Ok
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OpenStudy (anonymous):
time required to return to planet's surface
OpenStudy (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
OpenStudy (mathhelp346):
Ok
OpenStudy (mathhelp346):
So it's the same answer?
OpenStudy (anonymous):
no
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OpenStudy (anonymous):
wait this is some what lengthy
OpenStudy (anonymous):
now, time taken by the ball to reach the hand of the person is,
U/g + U/g = 2U/g
OpenStudy (mathhelp346):
Ok
OpenStudy (anonymous):
but the ball should reach the ground
OpenStudy (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
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OpenStudy (anonymous):
so it reaches the hands of the person with a velocity of 24 feet per second
OpenStudy (anonymous):
now it should reach the ground by traveling 3 feet
OpenStudy (mathhelp346):
So 27 feet
OpenStudy (anonymous):
no how do u say that?
OpenStudy (anonymous):
u der?
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OpenStudy (mathhelp346):
Yes
OpenStudy (anonymous):
now h(t) = (1/2)g(t^2) + Ut
3 = 16t^2 + 24t
OpenStudy (anonymous):
t = 0.116 seconds
OpenStudy (anonymous):
now add this with 2U/g
OpenStudy (anonymous):
hope u understood
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OpenStudy (mathhelp346):
There's a negative in front of (1/2)g
And how did you get 3?
OpenStudy (anonymous):
that is given because there the ball is projected vertically upward
here the ball is coming down
OpenStudy (anonymous):
it should travel 3 feet downward to reach the ground
OpenStudy (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?
OpenStudy (anonymous):
ya
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OpenStudy (mathhelp346):
So does that mean they all travel 3 feet downward?
OpenStudy (anonymous):
no
OpenStudy (mathhelp346):
but the ho for each of them is 3
OpenStudy (anonymous):
the ball should travel 3 feet downward after it reaches the hands of the person ,to reach the ground
OpenStudy (mathhelp346):
So how long do the other ones travel?
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