## Engineering Mechanics: Statics & Dynamics (14th Edition)

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.