A tall 2 meter tall basketball player shoots for the net that stands 3 meters from the ground. If he throws he ball from head height at an angle of 60 degrees and the ball travels at 10 meters per second, how far away is the hoop?

By taking the initial vertical speed as v_v = 6 sin(60) = 5.2 ms^-1.The max height of the ball is given by v_v² = u_v² + 2as. From this, s = 1.4m.Knowing the hieght, we can find the time that the ball spends in the air. To reach max height, use v = u + at to calculate t = 0.53. Then from max height, the ball falls a distance of 0.4m. To calculate how long it takes to fall this distance, use s = ut +0.5at² .As we now know the total time it takes for the ball to reach the hoop, we can calculate the horizontal distance the ball traveled.The horizontal speed, calculated by v = cos(60) = 3 ms^-1 can be used to measure the distance using simple time distance velocity formula, s = t/v = 2.46