Question about speed in physics.?

Question

Q: A penny is dropped from rest from the top of the building. Considering the height of the building is 427 meters & ignoring air resistance, find the speed with which the penny strikes the ground.

-I need help for my physics exam, any help would be greatly appreciated.

what if the answer must be in m/sec (squared)?
is it possible to solve it without using the conservation law? like by simply using sine, cosine or tangent?

Answer 1

The easiest way to solve this is to invoke the conservation of energy. Set h=0 to be the pavement/ground. You simply have the initial potential energy, mgh, equal to the final kinetic energy, (1/2)mv^2. That gives a speed of sqrt(2gh), about 91.5 m/s.

Answer 2

This is best found using an energy approach. The potential energy at the top of the building is m*g*h, or mass of penny*9.8*427. Then, at the bottom, the kinetic energy is mass*v^2/2. By conservation laws, these must be the same, so 9.8*427*2=v^2

Answer 3

The distance of fall is 427 m. Acceleration on Earth is 9.8 m / second sq. So at constant acceleration, the penny would hit the ground at 174 km per second.

For a more thorough explanation, go to
http://www.physicsclassroom.com/Class/1DKin/U1L1e.html

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Answer 4

Use the motion equation usually listed last:
Vf^2 = Vo^2 + 2*g*y

Answer 5

The penny hits the ground at 91.5 m/s. V=9.8t

Answer 6

I just don't see how you can ignore air resistance. There is such a thnig as terminal velocity. Maybe the links can help you.

Answer 7

Knowing that will be soooooo useful in your career..