qzwuf08sol

# Qzwuf08sol - PHY2054 Discussion-Fall 08 Warm Up Quiz(Physics 1 Review Name UFID*1(0pts A 1.80-m-tall basketball player stands on the floor 5.00 m

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August 29, 2008 PHY2054 Discussion-Fall ‘08 Warm Up Quiz (Physics 1 Review) Name: UFID: **1. (0pts) A 1.80-m-tall basketball player stands on the floor 5.00 m from the basket. If he/she shoots the ball at an angle of 45.0 ˚ to the horizontal, at what initial speed must he/she throw the basketball so that it goes through the hoop without striking the backboard? The height of the basket is 3.05 m. The time interval it takes the ball to reach the basket is expressed as Δx = v 0 cosθt t = Δx/(v 0 cosθ) When the ball reaches the basket, the vertical displacement should be Δy = 3.05-1.8 = 1.25 m. Plugging the expression for t into the kinematics equation for Δy, we solve the equation for v 0 : Δy = v 0 sinθt-(1/2)gt ² = v 0 sinθ[Δx/(v 0 cosθ)]-(1/2)g[Δx/(v 0 cosθ)] ² = Δxtanθ-(1/2)g[Δx/(v 0 cosθ)] 2 v 0 2 = gΔx ² /[2cos 2 θ(Δxtanθ-Δy)] v 0 = √[g/{2(Δxtanθ-Δy)}] × Δx/cosθ = √[9.8/{2(5tan45 ˚ -1.25)}] × 5/cos45 ˚ = 8.08 m/s ***2. (0pts) A 6.00-kg copper block on a 60.0

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## This note was uploaded on 12/13/2011 for the course PHY 2054 taught by Professor Avery during the Fall '08 term at University of Florida.

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Qzwuf08sol - PHY2054 Discussion-Fall 08 Warm Up Quiz(Physics 1 Review Name UFID*1(0pts A 1.80-m-tall basketball player stands on the floor 5.00 m

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