Step 1: Determine the effective acceleration of the ball relative to the lift. The lift is moving down with an acceleration $a_{lift} = 3 \text{ m/s}^2$. The acceleration due to gravity is $g = 9.8 \text{ m/s}^2$. Since both the ball (due to gravity) and the lift are accelerating downwards, the acceleration of the ball relative to the lift is the difference between their accelerations. $$a_{relative} = g - a_{lift}$$ $$a_{relative} = 9.8 \text{ m/s}^2 - 3 \text{ m/s}^2$$ $$a_{relative} = 6.8 \text{ m/s}^2$$ Step 2: Identify the initial conditions for the ball's motion relative to the lift. The ball is released, meaning its initial velocity relative to the lift is $u = 0 \text{ m/s}$. The distance the ball needs to fall to hit the floor is $s = 1.7 \text{ m}$. Step 3: Use a kinematic equation to find the time. We use the equation of motion: $$s = ut + \frac{1}{2}a_{relative}t^2$$ Since $u = 0$: $$s = \frac{1}{2}a_{relative}t^2$$ Rearrange the equation to solve for $t$: $$t^2 = \frac{2s}{a_{relative}}$$ $$t = \sqrt{\frac{2s}{a_{relative}}}$$ Step 4: Substitute the values and calculate the time. $$t = \sqrt{\frac{2 \times 1.7 \text{ m}}{6.8 \text{ m/s}^2}}$$ $$t = \sqrt{\frac{3.4 \text{ m}}{6.8 \text{ m/s}^2}}$$ $$t = \sqrt{0.5 \text{ s}^2}$$ $$t \approx 0.707 \text{ s}$$ Rounding to two significant figures: $$t \approx 0.71 \text{ s}$$ The ball will take approximately $\boxed{\text{0.71 s}}$ to hit the floor. Last free one today — make it count tomorrow, or type /upgrade for unlimited.