On the moon, the equation \( d=0.8 t^{2} \) is used to find the distance an object has fallen after \( t \) seconds. On Earth, the equation is \( d=4.9 t^{2} \). A rock is thrown from a height of 90 meters on Earth. Plot the graph of the equation on a graphing calculator or other technology to find the time taken to hit the ground to the nearest second.

To determine the time it takes for a rock to hit the ground when thrown from a height of 90 meters on Earth, we'll use the given equation for distance fallen under gravity: \[ d = 4.9 t^{2} \] Where: - \( d \) is the distance fallen in meters. - \( t \) is the time in seconds. **Step 1: Set Up the Equation** When the rock hits the ground, the distance fallen \( d \) will equal the initial height of 90 meters. Therefore: \[ 4.9 t^{2} = 90 \] **Step 2: Solve for \( t \)** 1. **Divide both sides by 4.9:** \[ t^{2} = \frac{90}{4.9} \approx 18.3673 \] 2. **Take the square root of both sides:** \[ t = \sqrt{18.3673} \approx 4.286 \text{ seconds} \] **Step 3: Round to the Nearest Second** \[ t \approx 4.286 \text{ seconds} \] Rounding to the nearest whole number: \[ t \approx 4 \text{ seconds} \] **Conclusion:** The rock will hit the ground approximately **4 seconds** after being thrown from a height of 90 meters on Earth. **Visual Representation:** If you plot the equation \( d = 4.9 t^{2} \) on a graphing calculator: 1. **Graph \( y = 4.9 t^{2} \).** 2. **Draw a horizontal line at \( y = 90 \) meters.** 3. **Find the intersection point of the two graphs.** This intersection occurs at \( t \approx 4.286 \) seconds, confirming our calculation. **Final Answer:** To the nearest second, the rock hits the ground after **4 seconds**.