A 47.0 g golf ball is driven from the tee with an initial speed of 50.0 m/s and

Question

A 47.0 g golf ball is driven from the tee with an initial speed of 50.0 m/s and rises to a height of 25.6 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. J (b) What is its speed when it is 10.0 m below its highest point?

Explanation / Answer

a)Maximum height = 25.6 Total Energy = constant =>Initial K.E. + Initial P.E. = Final K.E. + Final P.E. =>(0.5*(47/1000)*(50^2))+0 = (0.5*(47/1000)*v^2)+((47/1000)*9.81*25.6) =>v = 44.6959506 m/s b)Let the speed when it is 10 m below highest point be w. =>0.5*(47/1000)*w^2 = (0.5*(47/1000)*(44.6959506^2))+((47/1000)*9.81*10) =>w = 46.83938514 m/s

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