To solve Exercise 9, we need to determine how much longer the ball stays in the air on the Moon compared to Earth. We will calculate the total time the baseball spends in the air for both environments.

Step 1: Analyze the equations

1. Moon's equation: $h(t) = -0.8t^2 + 10t + 2$ Here, $h(t)$ represents the height of the ball at time $t$. The ball hits the ground when $h(t) = 0$.

2. Earth's equation: $h(t) = -4.9t^2 + 10t + 2$ Similarly, the ball hits the ground when $h(t) = 0$.

Step 2: Solve for $t$ when $h(t) = 0$ on the Moon

Set the Moon's equation to 0: $-0.8t^2 + 10t + 2 = 0$ Use the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = -0.8$
• $b = 10$
• $c = 2$

Substitute these values: $t = \frac{-10 \pm \sqrt{10^2 - 4(-0.8)(2)}}{2(-0.8)}$ $t = \frac{-10 \pm \sqrt{100 + 6.4}}{-1.6}$ $t = \frac{-10 \pm \sqrt{106.4}}{-1.6}$ $t = \frac{-10 \pm 10.32}{-1.6}$

Solve for the two roots: $t_1 = \frac{-10 + 10.32}{-1.6} = \frac{0.32}{-1.6} \approx -0.2 \, \text{(not valid, negative time is ignored)}$ $t_2 = \frac{-10 - 10.32}{-1.6} = \frac{-20.32}{-1.6} \approx 12.7 \, \text{seconds}$

Thus, the ball stays in the air for 12.7 seconds on the Moon.

Step 3: Solve for $t$ when $h(t) = 0$ on Earth

Set the Earth's equation to 0: $-4.9t^2 + 10t + 2 = 0$ Again, use the quadratic formula: $t = \frac{-10 \pm \sqrt{10^2 - 4(-4.9)(2)}}{2(-4.9)}$ Substitute the values: $t = \frac{-10 \pm \sqrt{100 - (-39.2)}}{-9.8}$ $t = \frac{-10 \pm \sqrt{139.2}}{-9.8}$ $t = \frac{-10 \pm 11.8}{-9.8}$

Solve for the two roots: $t_1 = \frac{-10 + 11.8}{-9.8} = \frac{1.8}{-9.8} \approx -0.18 \, \text{(not valid, negative time is ignored)}$ $t_2 = \frac{-10 - 11.8}{-9.8} = \frac{-21.8}{-9.8} \approx 2.22 \, \text{seconds}$

Thus, the ball stays in the air for 2.22 seconds on Earth.

Step 4: Compare the times

The ball stays in the air for:

• 12.7 seconds on the Moon.
• 2.22 seconds on Earth.

The difference in time is: $\Delta t = 12.7 - 2.22 \approx 10.48 \, \text{seconds}.$

Final Answer:

The ball stays in the air 10.48 seconds longer on the Moon than on Earth.

Would you like me to explain any part of this solution or go into more detail?

Follow-Up Questions:

1. What physical factors contribute to the difference in air time on the Moon versus Earth?
2. How does the gravitational acceleration on the Moon compare to Earth's?
3. Can you verify the calculations for each root using a calculator?
4. What happens if the initial velocity is changed on either planet?
5. How would air resistance on Earth alter the result?

Tip: Always ensure that your quadratic solutions make sense in the context of the problem (e.g., discard negative time values).