A 51.9-g golf ball is driven from the tee with an initial speed of 42.5 m/s and

Question

A 51.9-g golf ball is driven from the tee with an initial speed of 42.5 m/s and rises to a height of 28.3 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8.11 m below its highest point?

A 51.9-g golf ball is driven from the tee with an initial speed of 42.5 m/s and rises to a height of 28.3 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8.11 m below its highest point?

Explanation / Answer

a) using energy conservation,

initial energy = KE + PE = (m x 42.5^2 /2 ) + 0

final energy = ( mv^2 /2 ) + ( m x 9.81 x 28.3)

so, (m x 42.5^2 /2 ) + 0   = ( mv^2 /2 ) + ( m x 9.81 x 28.3)

42.5^2   = v^2 +   2 x 9.81 x 28.3

v = 35.37 m/s

b) then h = 28.3 - 8.11 = 20.19 m

using energy conservation,

(m x 42.5^2 /2 ) + 0   = ( mv^2 /2 ) + ( m x 9.81 x 20.19)

42.5^2   = v^2 +   2 x 9.81 x 20.19

v = 37.55 m/s

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