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Physics help

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ECAK
Posts: 150 - Karma: 50

A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop- ing under the tree. The constant speed of the horse is 12.2 m/s, and the woman is initially 1.6 m above the level of the saddle.

How long is she in the air? The acceleration of gravity is 9.8 m/s2 .

Answer in units of s


002 (part 2 of 2)

What must be the horizontal distance be- tween the saddle and limb when the woman makes her move?

Answer in units of m


any help would be great thanks. Sorry its kinda a last minute thing

Oct 29 2012 12:24AM
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TallerThanThou
Posts: 615 - Karma: 13
Find the time it would take the dumb crazy bitch to fall 1.6 meters (this should be easy because you know acceleration due to gravity and her initial velocity (which is zero)). Then find the distance the horse would travel in that amount of time. The distance the horse would travel is how far away it would need to be (assuming it instantaneously moves at 12 m/s at the exact moment the dumb crazy stunt bitch falls from the tree).
Oct 29 2012 12:29AM
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Wompuscat
Posts: 818 - Karma: 380
1st part - how long the woman is in the air.

You will use the equation x = Vot + .5at^2

what you know:
Vo = 0m/s
a = 9.8m/s^2
x = 1.6m

Unknown = t

Plug and use algebra and you get

1.6 = .5 x 9.8 x t^2

.3265 = t^2 take the squareroot of both sides and you get

.57 = t

she is in the air for .57 seconds.
Oct 29 2012 1:32AM
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Wompuscat
Posts: 818 - Karma: 380
2nd Part - Distance of horse

For this you will use the equation: X = .5(Vf-Vo)t

What you know:
Vf-Vo = average velocity = 12.2m/s
t=.57 seconds(how long the girl falls)

What you want to know:
X = Distance of horse from limb

Plug and use algebra:

X=.5 x 12.2 x .57

X = 3.477

The horse is 3.477 meters away from the limb.
Oct 29 2012 1:35AM
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Wompuscat
Posts: 818 - Karma: 380
thar ya be.
Oct 29 2012 1:35AM
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dreadyboy
Posts: 2306 - Karma: 34
then part 2 is just 0.57s x 12.2 m/s = 6.954m
Oct 29 2012 1:38AM
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Wompuscat
Posts: 818 - Karma: 380
yeah, my answer was completely wrong for the second part, whoops. I was just focusing on kinematics equations for some reason, my bad.
Oct 29 2012 1:53AM
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Stickity
Posts: 7750 - Karma: 769
SO EASY BRO
Oct 29 2012 2:52AM
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chicken
Posts: 6843 - Karma: 2,121
dude you have constant acceleration. so just integrate twice with respect to time to get position as a function of time. make sure to add the appropriate constants of integration (initial velocity=0, initial position =1.6) and then solve for t when d(t)=0

this gives you the formulas used above but if you realize where they come from it becomes super easy and you never have to memorize them.
Oct 29 2012 3:11AM
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dreadyboy
Posts: 2306 - Karma: 34
Yea, but if he can't figure this out, he prolly isn't in calculus and doesn't know what an integral is
Oct 29 2012 3:16AM
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chicken
Posts: 6843 - Karma: 2,121
when I said "dude you have constant acceleration" I didn't mean this will only work if you do. this method will work with changing acceleration as well.
Oct 29 2012 3:17AM
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