A basketball player throws the ball up to another player who dunks it at the hoop. The ball leaves the hands of the first player 6 feet above the ground at an upward velocity of 20 feet per second. The second player catches the ball right at the rim, which is 10 feet above the ground. How long does it take for the ball to get from the first players' hands to the second player.

The ball will take 1 sec

Step-by-step explanation:

∵ The distance from the ground when the ball leaves = 6 ft

∵ The distance from the ground when the ball catches = 10 ft

∴ The difference between two position = 10 - 6 = 4 ft

∵ s = ut - 1/2 at² ⇒ s is the distance, u is the initial velocity, a is acceleration of gravity and t is the time

∵ s = 4 feet, u = 20 feet/sec and a ≅ 32 feet/sec²

∴ 4 = 20t - 32/2 t²

∴ 16t² - 20t + 4 = 0

∴ (4t - 4) (4t - 1) = 0

∴ 4t - 4 = 0 ⇒ t = 1 and 4t - 1 = 0 ⇒ t = 1/4

∴ The ball will take 1 sec to go from first player to the second player

Note: 1/4 sec is the time of the ball after leaves the position of the first player by 4 feet above the initial position.