An athlete throws a basketball upward from the ground, giving it speed 9.7 m/s at an angle of 64.0° above the horizontal. (a) What is the acceleration of the basketball at the highest point in its trajectory? m/s2 (b) On its way down, the basketball hits the rim of the basket, 3.05 m above the floor. It bounces straight up with one-half the speed with which it hit the rim. What height above the floor does the basketball reach on this bounce? (a) is -9.8m/s2. For (b) I found the time when the ball hits the rim 1.3s. Then I found the final velocity -3.04m/s and divided by 2 to get -1.52m/s which is the initial velocity of the ball when it bounces straight up vertically from the rim. I am lost at where to go now and need help with solving (b).