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Physics 16 Online
OpenStudy (omnomnom):

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.

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

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.

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):

??

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?

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

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\[d =Vi( t)+ a \frac{ 1 }{ 2 }(t^2) \]

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?

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

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

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)

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

OpenStudy (anonymous):

http://www.physicsclassroom.com/

OpenStudy (anonymous):

this is a helpfull tool

OpenStudy (omnomnom):

oh kay thank you :D

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