A 47.0-g golfball is driven from the tee with an initial speed of 52.0 m/s and r

Question

A 47.0-g golfball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 24.6. (A) neglect air resistance and determine the KE of the ball at its highest point. (B) what is its speed when it is 8.0 m below its highest point? A 47.0-g golfball is driven from the tee with an initial speed of 52.0 m/s and rises to a height of 24.6. (A) neglect air resistance and determine the KE of the ball at its highest point. (B) what is its speed when it is 8.0 m below its highest point? (A) neglect air resistance and determine the KE of the ball at its highest point. (B) what is its speed when it is 8.0 m below its highest point?

Explanation / Answer

From the given question,

mass of ball (m)= 47g= 0.047kg

initial velocity of ball(u)= 52.0 m/s

maximum height(H)= 24.6m

A) Using conservation of energy

KE1 + PE1 = KE2 + PE2

(1/2) (0.047)(522) + (0.047)(10)(0) = KE2 + (0.047)(10)(24.6)

63.544 + 0 = KE2 + 11.562

KE2= 51.982

KE of ball at highest point is 52J

B) Height 8.0m below its highest point is 24.6-8=16.6m

Using conservation of energy

KE1 + PE1 = KE2 + PE2

(1/2) (0.047)(522) + (0.047)(10)(0) = (1/2)(0.047)(v2) + (0.047)(10)(16.6)

dividing both sides by (1/2)(0.047)

2704 + 0 = v2 + 332

v2 = 2372

v=48.7m/s

speed when 8.0m below its highest point is 48.7m/s

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