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A 47.0 g golf ball is driven from the tee with an initial speed of 54.0 m/s and rises to a height of 23.2 m. (a) Neglect air re...Asked by Brandon
A 47.0 g golf ball is driven from the tee with an initial speed of 54.0 m/s and rises to a height of 23.8 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?
m/s
(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?
m/s
Answers
Answered by
drwls
(a) Subtract the potential energy at 23.8 m from the initial kinetic energy.
(KE at highest pt.)
= (initial KE) - M g H
= (0.047 kg)*(54.0 m/s)^2/2
- (0.047)(9.8)(23.80 =
(b) 10.0 m below highest point is 13.8 m above tee elevation. Initial KE is decreased by M*g*13.8 m = 6.4 Joules at that elevation. Use that KE to get the corresponding speed
(KE at highest pt.)
= (initial KE) - M g H
= (0.047 kg)*(54.0 m/s)^2/2
- (0.047)(9.8)(23.80 =
(b) 10.0 m below highest point is 13.8 m above tee elevation. Initial KE is decreased by M*g*13.8 m = 6.4 Joules at that elevation. Use that KE to get the corresponding speed
Answered by
Physics2hard4me
This formula is not very clear to me
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