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Part 1) What is the momentum of the man as measured in the frame of the hot air balloon? p = kgm/s up the ladder

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Part 1) What is the momentum of the man as measured in the frame of the hot air balloon? p = kgm/s up the ladder Part 2) What is the velocity of the centre of mass of the man/hot air balloon system as the man climbs the ladder? vcm = m/s upwards Note: Enter a negative answer if it is downwards. Part 3) At the top of the ladder the man climbs into the basket. How far above the ground is the hot air balloon now? H: In Question: A hot air balloon is stationary H = 360 In above the ground with a man dangling from the bottom of a ladder which is L = 28.2 m long (Figure 1). The man begins to climb up the ladder with a speed 0 = 0.826 m/s relative to the ladder. The hot air balloon (excluding the man) has a mass of mbauoon = 252 kg, while the man has a mass mman = 87.5 kg. Figure 1: Diagrammatic representation of a man hanging off of the bottom of a ladder of length L, which extends upwards to the top ofthe balloon's basket. The man is climbing up at a speed of U m/s relative to the ladder. The balloon is at a height of H 111 when measured from the bottom ofthe balloon basket to the ground

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