Answer
The ball returns to its original position after 3.06 s.
Work Step by Step
1. Using this equation: $$s = s_0 + v_0t + \frac 12 at^2$$
- Since the ball will return to its original position: $s = s_0$
$$0 = v_0t + \frac 12 at^2$$
- The acceleration is going to be equal to the acceleration of gravity. Since it pulls the ball down, it must be negative.
$$0 = 15t + \frac 12 (-9.81)t^2$$
2. Solve for the time:
$$t_1 = \frac{-(15) + \sqrt{(15)^2 -4(\frac 12 (-9.81))(0)}}{2(\frac 12 (-9.81))} = 0 \space s$$
$$t_2 = \frac{-(15) - \sqrt{(15)^2 -4(\frac 12 (-9.81))(0)}}{2(\frac 12 (-9.81))} = 3.06 \space s$$
The ball will be at its start position at t = 0s and t = 3.06 s. Therefore, it returns to its original position after 3.06 s.