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Mathematics 57 Online
zarkam21:

?

zarkam21:

1 attachment
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

zarkam21:

so 9.8

Vocaloid:

yes

zarkam21:

x=0(1.75)+1/2(9.8)(1.75)^2

zarkam21:

15.0

Vocaloid:

good

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

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

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

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

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

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 =)

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

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