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August 29, 2008
PHY2054 DiscussionFall ‘08
Warm Up Quiz (Physics 1 Review)
Name:
UFID:
**1. (0pts) A 1.80mtall 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 = 3.051.8 = 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(ΔxtanθΔy)]
v
0
= √[g/{2(ΔxtanθΔy)}]
×
Δx/cosθ = √[9.8/{2(5tan45
˚
1.25)}]
×
5/cos45
˚
= 8.08 m/s
***2. (0pts) A 6.00kg 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.
 Fall '08
 Avery
 Physics

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