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\[mgh = \frac{ 1 }{ 2 }mv ^{2}\]

you can subtract the mass from both sides which shows that everything falls at the same rate

you know what g is

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Gravity?

yes

so if you plug in the velovity and solve for h which is the height

is this for conservation of energy

No this is only for the height of the building

The formula I gave you is the conservation of energy

In this case \(g\) is the gravitational acceleration we talked about in the previous problem.

You cannot solve this with kinematics there are too many unknowns

What do you mean you can't solve it with kinematics?

Yeah I guess you could solve for time in the acceleration then plug it in the average velocity.

I was talking about the \(g\) in your formula.

??

that is why I was asking

Last question was a 'kinematic' one so maybe it must be kinematics.

acceleration of gravity = velocity final minus velocity inital divided by time.

Like potential energy or kinetic energy?

No whats work?

It is something you will learn in the future, most likely.

yes solve for time then plug into \[\frac{ d }{ t }=\frac{ v _{f}+v _{i} }{ 2 }\]

conservation is so much easier to solve

it confused me though
the first equstion

equation*

Don't worry about it.

oh okay so what equation do i use.... is is\[d =Vi( t)+ a \frac{ 1 }{ 2 }(t^2) \]

You can't use that equation until you find time though.

Oh yeah >.<

\[a=\frac{ v _{f}-v _{i} }{ t }\]

you know a and both velocities

But we need the time :( Cant i just quess the answer?

Yes, \(a\) is the same as last time... \(9.81m/s^2\)

velocity initial is 0

You are not supposed to guess, because there is already a way to find the answer.

but...... Whats the way ?

t=49/9.81

answer is 5

yes 5 seconds

then plug in the time into your equation and find x

oh hold on

i got 49 m/s

that is the velocity

so we change the equation now?

you want the height of the building

yeah

x=1/2t(vfinal-vinital)

122.5 meters?

yes

thank you :D

do you understand the kinematic equations

The first one ? NO the others Yes

http://www.physicsclassroom.com/

this is a helpfull tool

oh kay thank you :D

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