physics
posted by anton .
You throw a 20N rock vertically into the air from ground level. You observe that when it is 15.0 m above the ground, it is travelling at 25.0 m/s upward. Use the workenergy theorem to find a) its speed just as it left the ground; b) its maximum height.

knowns:
F=20N
h=15m
V=25m/s
unknowns:
max h
V1
workenergy theorem: W=K2K1
K=.5*mV^2
so, m(g)h=.5mV2^2.5mV1^2
this simplifies to 2(g)h=V2^2V1^2
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 workenergy theorem again.
this time we get: 2(g)h=030.32^2
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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