An object is dropped from a tower, 1600 ft above the ground. The object's height above ground \( x \) seconds after the fall is \( s(x)=1600-16 \mathrm{x}^{2} \). About how long does it take the object to hit the ground? What is the object's velocity at the moment of impact? It takes \( \square \) sec for the object to hit the ground. (Round to the nearest tenth as needed.)

To solve the problem, we'll address both parts step by step. ### 1. Time to Hit the Ground **Given:** - Height function: \( s(x) = 1600 - 16x^2 \) - \( s(x) \) represents the height above the ground in feet. - \( x \) is the time in seconds. **To find:** The time \( x \) when the object hits the ground (\( s(x) = 0 \)). **Calculation:** \[ 1600 - 16x^2 = 0 \\ 16x^2 = 1600 \\ x^2 = \frac{1600}{16} \\ x^2 = 100 \\ x = \sqrt{100} \\ x = 10 \, \text{seconds} \] **Rounded to the nearest tenth:** \[ x = 10.0 \, \text{seconds} \] ### 2. Velocity at the Moment of Impact **Given:** - Height function: \( s(x) = 1600 - 16x^2 \) **To find:** The velocity \( v(x) \) at the moment of impact. **Calculation:** Velocity is the derivative of the position function with respect to time: \[ v(x) = \frac{ds}{dx} = \frac{d}{dx}(1600 - 16x^2) = -32x \] At \( x = 10 \) seconds (moment of impact): \[ v(10) = -32 \times 10 = -320 \, \text{ft/sec} \] **Interpretation:** - The negative sign indicates the direction of the velocity is downward. ### **Final Answer:** - **Time to hit the ground:** It takes **10.0 seconds** for the object to hit the ground. - **Velocity at impact:** The object's velocity at the moment of impact is **-320 ft/sec**.