Mathematics
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zarkam21:
?
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zarkam21:
Vocaloid:
|dw:1516753418724:dw|
Vocaloid:
find distance using equation 3, keeping in mind the initial velocity in the y-direction is 0
zarkam21:
x=0(1.75)+1/2(a)(1.75)^2
Vocaloid:
a is acceleration due to gravity
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zarkam21:
so 9.8
Vocaloid:
yes
zarkam21:
x=0(1.75)+1/2(9.8)(1.75)^2
zarkam21:
15.0
Vocaloid:
good
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zarkam21:
seconds?
Vocaloid:
oh god I used the wrong kinematics equation ughhh
zarkam21:
Its alright take your time
Vocaloid:
it needs to be equation 4 since we're looking for velocity
zarkam21:
okay
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Vocaloid:
anyway, initial is still 0, a is still g, and delta x is the height of the cliff
zarkam21:
V=0+2(9.8)(75)
Vocaloid:
v^2 not just v
zarkam21:
v^2=0+2(9.8)(75)
zarkam21:
38.34m
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Vocaloid:
what are the units of velocity?
zarkam21:
m/s
Vocaloid:
good so 38.4 m/s not m
Vocaloid:
anyway, you can use equation 3 to solve for time since you are given delta x, and a
zarkam21:
x=0(1.75)+1/2(9.8)(1.75)^2
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zarkam21:
wouldn't it be this for b
Vocaloid:
in this case we are solving for t so t is the unknown, plug everything else in and solve for t
zarkam21:
0(1.75)+1/2(9.8)t
zarkam21:
like this
Vocaloid:
it's t^2
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Vocaloid:
now set that equal to delta x then solve for t
zarkam21:
2.21
Vocaloid:
I got something a little different
deltax = 75 = (1/2)(9.8)t^2
t = ?
zarkam21:
3.91?
Vocaloid:
good, 3.91 seconds
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zarkam21:
wait this is b right?
zarkam21:
getting a little ahead of myself lol
Vocaloid:
yes
Vocaloid:
I just realized that our answer for part a) was actually the answer for c) not a since I used the height of the cliff ;;
zarkam21:
that's alright i just switched it over =)
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Vocaloid:
for a) you would actually just use equation 1, where vf = vi + at, t is the time given and vi = 0, a = 9.8
zarkam21:
Don't worry about it
zarkam21:
vf=0+(9.8)(1.75)
zarkam21:
17.15
Vocaloid:
yeah good, 17.15 m/s