Let's **estimate an upper bound for the height** of the space shuttle Endeavor **60 seconds after liftoff**. We are given the time and velocity data at several points. Since we're asked for an **upper bound**, we'll use the largest reasonable value for the velocity at each interval. We can estimate the height by integrating the velocity over time, i.e., finding the area under the velocity-time curve. Since we have only discrete data, we can **approximate the area under the curve using the velocities at each interval** (using the **right endpoint** for an upper bound). --- ### **Step 1: Write Down the Intervals and Velocities** | Interval (s) | Time Start | Time End | Velocity at End (ft/s) | Duration (s) | |--------------|------------|----------|------------------------|--------------| | 1 | 0 | 8 | 178 | 8 | | 2 | 8 | 15 | 338 | 7 | | 3 | 15 | 18 | 430 | 3 | | 4 | 18 | 30 | 754 | 12 | | 5 | 30 | 55 | 1,362 | 25 | | 6 | 55 | 60 | 1,461 | 5 | --- ### **Step 2: Calculate the Distance for Each Interval** We **approximate** distance traveled in each interval as: \[ \text{Distance} \approx \text{Velocity at End} \times \text{Duration} \] #### 1. Interval 0–8 s: \[ d_1 = 178 \ \text{ft/s} \times 8 \ \text{s} = 1,424 \ \text{ft} \] #### 2. Interval 8–15 s: \[ d_2 = 338 \ \text{ft/s} \times 7 \ \text{s} = 2,366 \ \text{ft} \] #### 3. Interval 15–18 s: \[ d_3 = 430 \ \text{ft/s} \times 3 \ \text{s} = 1,290 \ \text{ft} \] #### 4. Interval 18–30 s: \[ d_4 = 754 \ \text{ft/s} \times 12 \ \text{s} = 9,048 \ \text{ft} \] #### 5. Interval 30–55 s: \[ d_5 = 1,362 \ \text{ft/s} \times 25 \ \text{s} = 34,050 \ \text{ft} \] #### 6. Interval 55–60 s: \[ d_6 = 1,461 \ \text{ft/s} \times 5 \ \text{s} = 7,305 \ \text{ft} \] --- ### **Step 3: Sum All the Distances** \[ \text{Total Distance} = d_1 + d_2 + d_3 + d_4 + d_5 + d_6 \] \[ \text{Total Distance} = 1,424 + 2,366 + 1,290 + 9,048 + 34,050 + 7,305 \] \[ \text{Total Distance} = 55,483 \ \text{ft} \] --- ### **Step 4: Final Answer** **The upper bound for the space shuttle Endeavor’s height 60 seconds after liftoff is approximately:** \[ \boxed{55,483 \ \text{feet}} \] --- ### **Explanation** - We used the velocity at the *end* of each interval to get an upper bound, because the shuttle is accelerating (velocity is increasing). - This estimate **ignores air resistance and the fact that the shuttle is not going perfectly vertically**, but it's a reasonable upper estimate given the data. If you would like to see a lower bound or a more accurate estimate (average velocities), let me know!