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Date Posted: 7/24/2008 4:44:56 AM  Status: Live
PROBLEM
Course Textbook Chapter Problem
N/A N/A N/A N/A
Question Details:

A basketball player is fouled and knocked to the floor
during a lay-up attempt. The player is awarded two free-throws. The center of
the basket is a horizontal distance of 4.21m from the foul line and is a height
of 3.05m above the floor. On the first

free-throw attempt, he shoots the ball at an angle of 35°
above the horizontal and with a speed of

v0=4.88m/s. The ball is released 1.83m above the
floor. This shoot. misses badly. You can ignore  air resistance. 

a) What is the maximum height
reached by the floor?

b) at what distance along ?

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Date Posted: 7/24/2008 7:04:36 AM  Status: Live
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Response:
a)The maximum height reached by the floor is
h = [(vo2sin2 αo)/2g] --------(1)
Here,vo = 4.88m/s,αo = 35o and g = 9.8 m/s2.
Substituting the above values in equation (1),we get the value of maximum height.
b)The horizontal distance is
R = [(vo2 sin2αo)/g].



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