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3 April, 19:19

A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor at an angle of 60° above the horizontal, what must the initial speed be (in m/s) if it were to go through the basket?

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  1. 3 April, 22:02
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    Vi = 8.28 m/s

    Explanation:

    This problem is related to the projectile motion.

    As we know there are two components of motion associated with this, the horizontal component and vertical component.

    The horizontal distance covered by the ball is

    Vx*t = x

    Vx*t = 5.3

    Vx = 5.3/t eq. 1

    Also we know that

    Vx = Vicos (60)

    Vx = Vi*0.5 eq. 2

    equate eq. 1 and eq. 2

    5.3/t = Vi*0.5

    5.3/0.5 = Vi*t

    Vi*t = 10.6 eq. 3

    The vertical distance is

    Vy = y1 + Vyi*t - 0.5gt²

    also we know that

    Vyi = Visin (60)

    Vyi = Vi*0.866

    It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

    3 = 1.9 + Vi*0.866*t - 0.5gt²

    3 = 1.9 + Vi*0.866*t - 0.5 (9.8) t²

    3 = 1.9 + 0.866 (Vi*t) - 0.5 (9.8) t²

    3 = 1.9 + 0.866 (Vi*t) - 0.5 (9.8) t²

    1.1 = 0.866 (Vi*t) - 4.9t²

    0.866 (Vi*t) = 4.9t² + 1.1

    substitute Vi*t = 10.6 in above equation

    0.866 (10.6) = 4.9t² + 1.1

    9.18 = 4.9t² + 1.1

    4.9t² = 8.08

    t² = 8.08/4.9

    t² = 1.648

    t = 1.28 sec

    Finally, initial speed can be found by substituting the value of t into eq. 3

    Vi*t = 10.6

    Vi = 10.6/t

    Vi = 10.6/1.28

    Vi = 8.28 m/s
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