I just need to reassurance, would they both be negative?
2 answers:
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Answer:
So energy is not conserved and inelastic shock
Explanation:
In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.
Let's find the kinetic energy just before the crash
K₀ = ½ m vₓ₀²
After the crash we can use the law of conservation of energy
Starting point. Right after the crash, before starting to climb
Em₀ = K = ½ (m + M) v₂²
Final point. At the maximum height of the pendulum
= U = (m + M) g h
Where m is the mass of the bullet and M is the mass of the pendulum
Em₀ = Em_{f}
½ (m + M) v₂² = (m + M) g h
v₂ = √ 2g h
Now we can calculate the final kinetic energy
K_{f} = ½ (m + M) v₂²²
K_{f} = ½ (m + M) (2gh)
The relationship between these two kinetic energies is
K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)
K₀ / K_{f} = m / (m + M) vₓ₀² / 2 g h
We can see that in this relationship the Ko> Kf
So energy is not conserved and inelastic shock
The answer would be 371.16 ft.
Refer to the diagram for the scenario. The drawing is not to scale.
Notice the diagram that you will find to adjacent right triangles. All you need to do is first solve for the needed side of one triangle, by using one side.
It sounds confusing I know, but let's do this step by step.
Angle of depression is 15°.
We use this first because we do not have enough data to use the angle of elevation.
To get the adjacent side, we use the trigonometric function TOA which means:
We use the height of the first building as our reference, so our given will be:
Opposite = 20ft
θ = 15°
Now we put that into our formula:
Angle of elevation is 78°:
Now that we have that we can solve for the height from the point of 20ft. We use the same formula, because we are looking for the opposite, given the adjacent.
Now that's the height from the top of the first building. To get the total height of the building we just add the 20ft.
351.16ft + 20 ft = 371.16ft
<em>The building across the street is 371.16 ft tall.</em>
