Question
You throw a 20-N 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 work-energy theorem to find a) its speed just as it left the ground; b) its maximum height.
Answers
knowns:
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
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
Thank you
Ook