A girl rolls a ball up an incline and allows it to return to her. For the angle! and ball involved, the acceleration of the ball along the incline is constant at 0.25g, directed down the incline. If the ball is released with a speed of 4 m / s, determine the distance s it moves up the incline before reversing its direction and the total time t required for the ball to return to the child's hand.

The distance the ball moves up the incline before reversing its direction is 3.2653 m.

The total time required for the ball to return to the child's hand is 3.2654 s.

Explanation:

When the girl is moving up:

The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = - 0.25g = - 0.25*9.8 = - 2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t to reach the top.

Using

v = u + a*t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

Total time to go and come back = 2 * 1.6327 = 3.2654 s

To find the distance, using:

v² = u² + 2*a*s

0² = 4² + 2 * (-2.45) * s

s = 3.2653 m

Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.