You are designing a slide for a water park. In a sitting position, park guests slide a vertical distance h down the waterslide, which has negligible friction. When they reach the bottom of the slide, they grab a handle at the bottom end of a 6.00mlong uniform pole. The pole hangs vertically, initially at rest. The upper end of the pole is pivoted about a stationary, frictionless axle. The pole with a person hanging on the end swings up through an angle of 72.0∘, and then the person lets go of the pole and drops into a pool of water. Treat the person as a point mass. The pole's moment of inertia is given by I=13ML2, where L = 6.00 m is the length of the pole and M = 31.0 kg is its mass.For a person of mass 70.0 kg, what must be the height h in order for the pole to have a maximum angle of swing of 72.0∘ after the collision?


h = 
35.835.8mm 
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When Person started the slide it's energy
= M_{person}gh = 70*9.81*h = 686.7h (1)
When it reaches at the bottom of the pole this potential energy
must be converted in to the kinetic energy
(1/2)M_{person}*V^{2} = 686.7h (by conservation of
energy)
Now this will converted into the potential energy when it reaches
at the extreme position as shown in the figure.
Potential energy = M_{P}gH + M_{Rod}*gP
Where H = L  Lcos72 = 6  6Cos72 = 4.146 m
P = L  (L/2)Cos72 = 5.073 m
Potential energy = (70*9.81*4.146) + (31*9.81*5.073) = 4389.808
J
Hence by energy conservation
686.7h = 4389.808
h = 6.393 m
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