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March 1, 2015

March 1, 2015

Posted by **anton** on Sunday, February 21, 2010 at 2:39am.

- physics -
**Timmy**, Thursday, March 10, 2011 at 3:25amknowns:

F=20N

h=15m

V=25m/s

unknowns:

max h

V1

work-energy theorem: W=K2-K1

K=.5*mV^2

so, m(-g)h=.5mV2^2-.5mV1^2

this simplifies to 2(-g)h=V2^2-V1^2

therefore, V1=sqrt(V2^2+2gh)

now all you have to do is plug in numbers.

V1=sqrt(25^2+2*9.8*15)=30.32m/s

max height is where V=0

we can use the work-energy theorem again.

this time we get: 2(-g)h=0-30.32^2

so, max h=-(30.32^2)/2(-g)=46.90m

the final answer is:

max height = 46.90m

and initial velocity = 30.32m/s

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