Question

you throw a 20n rock vertically into the air from ground. you observe that when it is 15m above the grount it is travelling at a velocity of 25m/s. use work energy theorem to find its speed when it left the ground and its maximum height

Answer 1

Answer:

ground; b) its maximum height.

anton

Feb 21, 2010

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

Explanation:

please drop some thanks for me

Answer 2

Explanation:

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