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posted by  Learning on 6/11/2007 3:18:46 PM  |  status: Live  

Conservation of Mechanical Energy: Cutnell 7th Ed.Ch.6 #37

Course Textbook Chapter Problem
Algebra Based Physics Physics 7th Ed. Cutnell 6 37
Question Details:
37.  A 47.0 g golf ball is driven from the tee with an initial speed -52.0 m/s and rises to a height of 24.6 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 8.0m below its highest point?
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posted by HungryHungryHippo on 6/11/2007 4:00:54 PM  |  status: Live
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Response Details:
(a) Use conservation of energy.  The kinetic energy when the ball leaves the tee (PE = 0) will equal the total energy of the ball at the top of its path (KE + PE):

KEi = KEf + PEf
1/2 mvi2 = 1/2 mvf2 + mgh 

(reduces to...)

vi2 = vf2 + 2gh
(-52)2 = vf2 + 2(9.8)(24.6)
vf2 = 2222 m2/s2
vf = 47.1 m/s

(b) Do the same thing as part (a), except for a height 8.0 m below its highest point

KEi = KEf + PEf
1/2 mvi2 = 1/2 mvf2 + mgh
vi2 = vf2 + 2gh
(-52)2 = vf2 + 2(9.8)(24.6 - 8.0)
vf2 = 2379 m2/s2
vf = 48.8 m/s
Hope I have helped.  Anyone feel free to correct any mistake I may have made.
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