document.write( "Question 1194654: A golfer hits his approach shot at an angle of 50° giving the ball an intial speed of 32.2 m/s. The ball lands on an elevated green,5.5 above the intial position near the surface,and stops immediately.
\n" ); document.write( " a) how much time passed while the ball was in air?
\n" ); document.write( " B) how far did the ball travel horizontally before landing?
\n" ); document.write( " c) what was the peak height reached by the ball?
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Algebra.Com's Answer #827195 by Alan3354(69443)\"\" \"About 
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A golfer hits his approach shot at an angle of 50° giving the ball an intial speed of 32.2 m/s. The ball lands on an elevated green,5.5 above the intial position near the surface,and stops immediately.
\n" ); document.write( " a) how much time passed while the ball was in air?
\n" ); document.write( " B) how far did the ball travel horizontally before landing?
\n" ); document.write( " c) what was the peak height reached by the ball?
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\n" ); document.write( "Using 10 m/sec/sec for gravity and no air resistance:
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\n" ); document.write( "The vertical speed at launch is 32.2*sin(50)
\n" ); document.write( "The time to max height is 32.2*sin(50)/10 = 3.22*sin(50) seconds
\n" ); document.write( "c) Max height = gt^2/2 = 5*(3.22sin(50))^2 = ~ 30.422 meters
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\n" ); document.write( "h(t) = 32.2sin(50)t - 5t^2 = 5.5 meters
\n" ); document.write( "5t^2 - 32.2sin(50)t + 5.5 = 0
\n" ); document.write( "t = 0.234 seconds ascending thru 5.5 meters
\n" ); document.write( "t = 4.6992 seconds at impact at h = 5.5 meters
\n" ); document.write( "a) how much time passed while the ball was in air?
\n" ); document.write( "~ 4.6992 seconds
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\n" ); document.write( "Horizontal component of speed is 32.2*cos(50) =~ 20.698 m/sec
\n" ); document.write( "b) how far did the ball travel horizontally before landing?
\n" ); document.write( "= 20.698*4.6992 = 97.26 meters
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\n" ); document.write( "PS Golf is boring
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