A stone is dropped from the roof of a tall building. A person measures the speed of the stone to be 49 m/sec when it hits the ground. The height of the building is closest to:
Select one:
a. 24 meters.
b. 49 meters.
c. 122 meters.
d. 245 meters.
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OpenStudy (anonymous):
\[mgh = \frac{ 1 }{ 2 }mv ^{2}\]
OpenStudy (anonymous):
you can subtract the mass from both sides which shows that everything falls at the same rate
OpenStudy (anonymous):
you know what g is
OpenStudy (omnomnom):
Gravity?
OpenStudy (anonymous):
yes
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OpenStudy (anonymous):
so if you plug in the velovity and solve for h which is the height
OpenStudy (anonymous):
is this for conservation of energy
OpenStudy (omnomnom):
No this is only for the height of the building
OpenStudy (anonymous):
The formula I gave you is the conservation of energy
OpenStudy (anonymous):
In this case \(g\) is the gravitational acceleration we talked about in the previous problem.
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OpenStudy (anonymous):
You cannot solve this with kinematics there are too many unknowns
OpenStudy (anonymous):
What do you mean you can't solve it with kinematics?
OpenStudy (anonymous):
Yeah I guess you could solve for time in the acceleration then plug it in the average velocity.
OpenStudy (anonymous):
I was talking about the \(g\) in your formula.
OpenStudy (omnomnom):
??
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OpenStudy (anonymous):
Though technically speaking, depending on whether they are on the 'energy' part or the 'kinematic' part of the course decides which method they should use.
OpenStudy (anonymous):
that is why I was asking
OpenStudy (anonymous):
Last question was a 'kinematic' one so maybe it must be kinematics.
OpenStudy (anonymous):
acceleration of gravity = velocity final minus velocity inital divided by time.
OpenStudy (anonymous):
@omnomnom Have you talked about work or energy in your class yet?
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OpenStudy (anonymous):
Like potential energy or kinetic energy?
OpenStudy (omnomnom):
No whats work?
OpenStudy (anonymous):
It is something you will learn in the future, most likely.
OpenStudy (anonymous):
yes solve for time then plug into \[\frac{ d }{ t }=\frac{ v _{f}+v _{i} }{ 2 }\]
OpenStudy (anonymous):
conservation is so much easier to solve
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OpenStudy (omnomnom):
it confused me though
the first equstion
OpenStudy (omnomnom):
equation*
OpenStudy (anonymous):
Don't worry about it.
OpenStudy (anonymous):
@omnomnom This is a two step problem. First you want to figure out how long it took to fall. Then you want to use that to figure out how far it fell.
OpenStudy (omnomnom):
oh okay so what equation do i use.... is is\[d =Vi( t)+ a \frac{ 1 }{ 2 }(t^2) \]
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OpenStudy (anonymous):
You can't use that equation until you find time though.
OpenStudy (omnomnom):
Oh yeah >.<
OpenStudy (anonymous):
\[a=\frac{ v _{f}-v _{i} }{ t }\]
OpenStudy (anonymous):
you know a and both velocities
OpenStudy (omnomnom):
But we need the time :( Cant i just quess the answer?
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OpenStudy (anonymous):
Yes, \(a\) is the same as last time... \(9.81m/s^2\)
OpenStudy (anonymous):
velocity initial is 0
OpenStudy (anonymous):
You are not supposed to guess, because there is already a way to find the answer.
OpenStudy (omnomnom):
but...... Whats the way ?
OpenStudy (anonymous):
t=49/9.81
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OpenStudy (omnomnom):
answer is 5
OpenStudy (anonymous):
yes 5 seconds
OpenStudy (anonymous):
then plug in the time into your equation and find x
OpenStudy (omnomnom):
oh hold on
OpenStudy (omnomnom):
i got 49 m/s
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OpenStudy (anonymous):
that is the velocity
OpenStudy (omnomnom):
so we change the equation now?
OpenStudy (anonymous):
you want the height of the building
OpenStudy (omnomnom):
yeah
OpenStudy (anonymous):
x=1/2t(vfinal-vinital)
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OpenStudy (omnomnom):
122.5 meters?
OpenStudy (anonymous):
yes
OpenStudy (omnomnom):
thank you :D
OpenStudy (anonymous):
do you understand the kinematic equations
OpenStudy (omnomnom):
The first one ? NO the others Yes
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