Question

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.

Answer 1

To determine the distance and the total time for the ball to return to the girl, we apply kinematic equations with the initial velocity of 4 m/s and a constant acceleration of 0.25g down the incline. The ball travels 3.27 meters up the incline before stopping, and the total time for the round trip is 3.26 seconds.

Step-by-step explanation:

To determine the distance s that a ball moves up an incline before reversing its direction and the total time t required for it to return to the girl, we can use kinematics equations. The acceleration a is given as 0.25g, which means the acceleration is 0.25 times the acceleration due to gravity, with g being 9.8 m/s². The initial velocity u of the ball is 4 m/s up the incline.

The final velocity v when the ball stops at the highest point is 0 m/s. By using the kinematic equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can solve for s:

1. v² = u² + 2as
2. 0 = (4 m/s)² - 2(0.25 * 9.8 m/s² * s)
3. s = (4 m/s)² / (2 * 0.25 * 9.8 m/s²)
4. s = 16 m²/s² / 4.9 m/s²
5. s = 3.27 m

The ball travels a distance of s = 3.27 m up the incline before stopping and reversing its direction.

To find the total time t, we know that the time taken to go up and down the incline is the same. We use the equation v = u + at, and by rearranging for t, we get t = (v - u) / a.

1. t = (0 - 4 m/s) / (-0.25 * 9.8 m/s²)
2. t = -4 m/s / -2.45 m/s²
3. t = 1.63 s

The ball takes 1.63 seconds to reach the highest point, so the total time for the round trip is t = 1.63 s * 2 = 3.26 s. Thus, it takes 3.26 seconds for the ball to return to the girl's hand.

Answer 2

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.

Step-by-step 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.