Equilibrium and impulsive forces (crane and demolition ball problem)

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Homework Statement


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The Attempt at a Solution


I tried to find the acceleration using F= ma

Resultant force = Tension - mg
so a = (10295.2-8730)/890 = 1.758 m/s^2 but I don't think it works that way...

please help.
 
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The height which from the ball falls, and its mass, gives a potential energy to the ball. This results in an identical kinetic energy at 0-level. The angle and length of rope should give the height. Then impulse can be calculated.
 
I don't think so, because at the vertical position the total energy would equal kinetic energy + some potential energy therefore mgh = 0.5 mv^2 + mgh

And both velocity & height are missing from the right side of the equation.
 
I think the height can be calculated by angle and length, but possibly that is not what is asked for. My reply might be uninformative. Tension is not a topic of mine. Nor is geometry, really.
 
Can you please attempt it? I have solved other parts of this question and I will see if your calculated height fits in with the rest.
 
At vertical position, the ball can fall no further in this system: equals no potential energy
 
I tried it and the final velocity came up as 17 m/s, I know this is wrong because in the next part it says SHOW that the final velocity just before impact = 6.7 m/s
 
This is ridiculous... I'm sure the answer should be fairly easy.
 
The 32deg. form the top of a triangle. The two other angles are 180-32=148 deg. combined, or 74 deg. each. Find length of base of triangle. Probably some cosinus or sinus calculating. I don't know. Then there's another small triangle below, whith 90 and 90-74=16 deg. angles. Base from other triangle is one side, short side of this small triangle is your heigth, and basis for potential/kinetic energy
 
Thank you but I still think there is some other solution that jiggles with Newton's laws. If someone figures it please post.
 
has anyone got any idea, please? At this point I tried everything and I just can't get my answer to fit both a final velocity of 6.7 m/s and a reasonable height
 
I figured this and thought I'd post the solution:

Height at elevated position = 15 cos 32 = 12.72m

Therefore h = 15-12.72 = 2.28m