A 47.0 g golf ball is driven from the tee with an initial speed of 56.0 m/s and

Question

A 47.0 g golf ball is driven from the tee with an initial speed of 56.0 m/s and rises to a height of 24.4 m. A) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (Answer in J) B) What is its speed when it is 9.0 m below its highest point? (Answer in m/s)

Explanation / Answer

( a ) By the law of conservation of mechanical energy, Initial kinetic + potential energy = Final kinetic + potential energy => (1/2) mu^2 + 0 = kinetic energy at the highest point + mgh => kinetic energy at the highest point = (1/2) mu^2 - mgh = (0.047) * [(1/2) (56)^2 - (9.81) * (24.2)] J = 62.54 J. ( b ) Let v = speed at 7 m below the highest point Kinetic energy at 7 m below the highest point = kinetic energy at the highest point - loss of potential energy => (1/2) mv^2 = 62.54 - (0.047) * (9.81) * (7.0) => (1/2) mv^2 = 59.31 J => v = v[(59.31) * 2 /(0.047)] = 50.24 m/s

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