SOLUTION: 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,
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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.
a) how much time passed while the ball was in air?
B) how far did the ball travel horizontally before landing?
c) what was the peak height reached by the ball? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! 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.
a) how much time passed while the ball was in air?
B) how far did the ball travel horizontally before landing?
c) what was the peak height reached by the ball?
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Using 10 m/sec/sec for gravity and no air resistance:
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The vertical speed at launch is 32.2*sin(50)
The time to max height is 32.2*sin(50)/10 = 3.22*sin(50) seconds
c) Max height = gt^2/2 = 5*(3.22sin(50))^2 = ~ 30.422 meters
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h(t) = 32.2sin(50)t - 5t^2 = 5.5 meters
5t^2 - 32.2sin(50)t + 5.5 = 0
t = 0.234 seconds ascending thru 5.5 meters
t = 4.6992 seconds at impact at h = 5.5 meters
a) how much time passed while the ball was in air?
~ 4.6992 seconds
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Horizontal component of speed is 32.2*cos(50) =~ 20.698 m/sec
b) how far did the ball travel horizontally before landing?
= 20.698*4.6992 = 97.26 meters
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PS Golf is boring
