a basketball is shot from an initial height of 2.4 m with an initial speed vo= 9

Question

a basketball is shot from an initial height of 2.4 m with an initial speed vo= 9 m/s directed at an angle theta 0=35 degrees above the horizontal? Height of the basket is 3.05 m At what angle to the horizontal did the ball enter the basket? a basketball is shot from an initial height of 2.4 m with an initial speed vo= 9 m/s directed at an angle theta 0=35 degrees above the horizontal? Height of the basket is 3.05 m At what angle to the horizontal did the ball enter the basket? a basketball is shot from an initial height of 2.4 m with an initial speed vo= 9 m/s directed at an angle theta 0=35 degrees above the horizontal? Height of the basket is 3.05 m At what angle to the horizontal did the ball enter the basket? a basketball is shot from an initial height of 2.4 m with an initial speed vo= 9 m/s directed at an angle theta 0=35 degrees above the horizontal? Height of the basket is 3.05 m At what angle to the horizontal did the ball enter the basket?

Explanation / Answer

height of basket from initial point = 3.05 - 2.4 = 0.65 m

horizontal velocity = 9cos35 = 7.37237 m/s

initial vertical velocity = 9sin35 = 5.1622 m/s

vertical velocity at basket height = sqrt(5.16222 - 2 x 9.8 x 0.65) = 3.7294 m/s

angle to the horizontal = tan-1(3.7294/7.37237) = 26.833o (below the horizontal)

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