A basketball player shoots a 0.75-kg ball into a basket byshooting the ball with

Question

A basketball player shoots a 0.75-kg ball into a basket byshooting the ball with an initial speed of 12 m/s at an angle of 60 above the horizontal. The basket is threemeters o the ground and the ball leaves her hand at a height of 1.4 meters above the ground. a) How far away from directly under the basket is the playerstanding? b) What is the velocity of the ball when it goes into thebasket? (In vector notation form) A basketball player shoots a 0.75-kg ball into a basket byshooting the ball with an initial speed of 12 m/s at an angle of 60 above the horizontal. The basket is threemeters o the ground and the ball leaves her hand at a height of 1.4 meters above the ground. a) How far away from directly under the basket is the playerstanding? b) What is the velocity of the ball when it goes into thebasket? (In vector notation form)

Explanation / Answer

v = 12 m/s, = 60o, vertical displacement y = 3 - 1.4 = 1.6 m initial velocity vx = vcos = 6 m/s, vy= vsin = 10.392 m/s y = vyt - gt2/2 4.9t2 - 10.392t + 1.6 = 0 t = 0.167 s a) x = vxt = 1.00 m b) vx' = vx = 6.0 m/s vy' = vy - gt = 8.76 m/s answer: (6.0 i + 8.76 j) m/s

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