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Physics: 2D Kinematics

Posted on 2011-09-21
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Last Modified: 2012-05-12
A rock is thrown horizontally at a speed of 80ft/sec from the top of a building that is 64 ft high.

A) How long does it take for the rock to hit the land?

B) How far from the base of the building will the rock land?

C) At what speed does the rock hit the land?

My answers so far are:
a) t = 1.995 seconds to hit ground
b) rock will land 48.68 meters (or 159.62 feet) away from the base of the building
c) Haven't finished solving but I keep getting v^2 = -358.68 m/s which is impossible... Unless it means 358 m/s in the opposite direction, which DOES make sense, since velocity is a vector which has direction and magnitude.
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Question by:Zenoture
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by:ozo
ID: 36578592
with what vertical velocity will the rock hit the land?
with what horizontal velocity will the rock hit the land?
or
with what kinetic energy will the rock hit the land?
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Author Comment

by:Zenoture
ID: 36578638
Don't think I'm allowed to use KE since this is for my multivariable calc class, but vert. velocity is v*sin(@), and horizontal velocity is v*cos(@), but we know vo = <24.4, 0> making our vo[x] = 24.4
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Expert Comment

by:ozo
ID: 36578703
How did you get v^2 = -358.68 m/s?
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Accepted Solution

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Zenoture earned 0 total points
ID: 36578715
v^2 = vo^2 - 2gd

But after revising what I had and thinking simple, I realized that Vx is constant, since ax = 0, therefore Vfx = Vox = 24.4 m/s

Vfy = Voy - gt = 0 - (9.8)*(1.99) = -19.50 j

|| Vf || = sqrt( Vfx^2 + Vfy^2 ) = 31.23 m/s
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Assisted Solution

by:ozo
ozo earned 500 total points
ID: 36578727
That makes more sense than sqrt(-358.68 m/s)
which would have units of i*sqrt(m)/sqrt(s)

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Author Closing Comment

by:Zenoture
ID: 36597844
The full work out is shown in my answer, but was arrived at by the help of ozo.
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