You roll a ball up a hill, releasing the ball at 12 m/s. The angle of the hill gives the ball an acceleration of 6.0 m/s/s, directed down the hill. How long will it take the ball to return to you at the bottom of the hill

it take the ball to return to you at 4 sec

Explanation:

given data

releasing the ball = 12 m/s

acceleration a = 6.0 m/s²

solution

when releasing the ball is 12m/s

so initial speed will be  u = 12m/s  

and final speed will be  v = -12m/s  

and when ball is rolled up with 12 m/s so by by law of conservation of energy ball roll back to same point with same speed  so we use here  kinematic equation  

v = u + at    1

put here value and we get

-12 = 12 + (-6) × t  

solve it we get time t

time = 4 sec

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

need to see circuit to get time constant