- #1
fara0815
- 45
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Hello there !
I have been trying to figure the following problem out but unfortunately I have not had any obvious success. It is about Kinetic Energy:
A stone is falling from the roof of a house down to the ground. In the last second of the fall, the stone's energy increases of 293 J.
How high is the house?
(friction is neglected)
I have gone several ways but never got the stated result of 20,06 m.
I would be very happy if anyone could give me a hint to solve this one. It seems so easy but I cannot make it :(
The most logical way - at least for me - would be: (the ground is zero in my coordinate system)
∆Ekin= 293 J = 1/2 * m * v^2 thus: V= - 17,13 m/s
V is the velocity on average in the last second. Thus the calculated velocity is actually there at tf-0,5s. With Tf is meant the over all time of the fall.
V= - g*t, thus t= 1.745 s
tf= t + 0.5 s = 2,245 s
h= 1/2*g*t^2= 24,72 m
Any suggestions ?
I would be really happy to get some help with this one !
I have been trying to figure the following problem out but unfortunately I have not had any obvious success. It is about Kinetic Energy:
A stone is falling from the roof of a house down to the ground. In the last second of the fall, the stone's energy increases of 293 J.
How high is the house?
(friction is neglected)
I have gone several ways but never got the stated result of 20,06 m.
I would be very happy if anyone could give me a hint to solve this one. It seems so easy but I cannot make it :(
The most logical way - at least for me - would be: (the ground is zero in my coordinate system)
∆Ekin= 293 J = 1/2 * m * v^2 thus: V= - 17,13 m/s
V is the velocity on average in the last second. Thus the calculated velocity is actually there at tf-0,5s. With Tf is meant the over all time of the fall.
V= - g*t, thus t= 1.745 s
tf= t + 0.5 s = 2,245 s
h= 1/2*g*t^2= 24,72 m
Any suggestions ?
I would be really happy to get some help with this one !
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