Asked by Savegg
                A 2.0 m-tall basketball player is standing on the floor 4.0 m from the basket. If he shoots the ball at a 52.0° angle with the horizontal, at what initial speed does the ball just barely goes through the rim neatly if the height of the basket is 3.05 m and the diameters of the basketball and the rim are 24 cm and 46 cm ?
            
            
        Answers
                    Answered by
            henry2, 
            
    Range = Vo^2*sin(2A)/g = 4.
Vo^2*sin(104)/9.8 = 4
Vo^2*0.0990 = 4
Vo = 6.4m/s[52o].
    
Vo^2*sin(104)/9.8 = 4
Vo^2*0.0990 = 4
Vo = 6.4m/s[52o].
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