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EX9.1-Golf_Shot-aud-2

# EX9.1-Golf_Shot-aud-2 - x2 mv x1 R y ∆ t = mv y2-mv y1 R...

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Lesson 9 Example 9.1 Golf Shot Hi-speed Video of Golf Ball

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3/216 The 1.62 oz golf ball is struck with a five iron and acquires a velocity of 150 ft/sec in a time period of 0.001 sec. Determine the magnitude R of the average force exerted by the club on the ball. What is your first step? EX9.1 Golf Shot Solution: Problem Type: Find: Given: W ball =1.62 oz, v 2 =150 ft/s, Δt=0.001 sec R, average force exerted on the ball Kinetics: Impulse-Momentum Reflection: Note that the force is much greater than the weight of the ball so our assumption that we can neglect the ball’s weight seems like a good one. Draw impulse-momentum diagrams R x t = mv
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Unformatted text preview: x2 - mv x1 R y ∆ t = mv y2-mv y1 R x = mv x2 / ∆ t R y = mv y2 / ∆ t ( 29 001 . / 25 cos 150 2 . 32 16 / 62 . 1 = x R ( 29 001 . / 25 sin 150 2 . 32 16 / 62 . 1 = y R R x = 427.5 lb R y = 199.3 lb Solve R = 471.7 lbs Could also set our axes along the direction of v-∫ 2 1 t t dt F R 1 2 v m v m R R-= Most of the time the impact forces are much greater than the weight of the object, so we can neglect mg. You can either break it into components for the impulse momentum equation, or break it up later – let’s start by writing equations in the x and y. x y mv x = mv y = 0 + 2 v m R = ∆ t...
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