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Let g be the acceleration of gravity near the Earth's surface
Let g be the acceleration of gravity near the Earth's surface. The acceleration of gravity near the surface of the Moon is (approximately) g/6. Using the law of conservation of energy, i.e. the principle of conservation of energy, solve the following.
(a) Suppose that a ball is dropped from 9 feet above the Earth.
Taking g = 32 feet/second square, at what speed is the ball traveling as it reaches the Earth?
(b) Suppose that a ball is drpped from 9 feet above the Moon. At what speed is the ball traveling as it reaches the Moon?
(c) Suppose that a ball dropped from height he above the Earth's surface strikes the ground with the same speed as a ball dropped from a height hm above the Moon's surface. Calculate hm/he.
Expert Solution
Please see the solution in the attached word file 'Solution_Bouncing_ball_01_by_EnergyConservation.doc'
The soliution of this problem is very simple. Throughout, you have to note that total energy of the ball remains conserved.
Total energy = Kinetic energy + Potential Energy
Hence constant total energy of the ball means
KE + PE = constant
With the position of the ball, its potential energy (PE) changes and hence its velocity too. Therefore its Kinetic Energy (KE) also keeps on changing.
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