Step-by-Step Solution

Step-by-Step Solution

Step 1: Set up the altitude inequality

The height must be at least $40\text{ m}$: $-5t^2 + 20t + 30 \ge 40$

$-5t^2 + 20t - 10 \ge 0$

Divide by $-5$ (reversing the inequality sign from $\ge$ to $\le$): $t^2 - 4t + 2 \le 0$

Step 2: Find boundary entry and exit times

Solve the boundary quadratic equation $t^2 - 4t + 2 = 0$: $t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$

$t = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2}$

Simplify the radical ($\sqrt{8} = 2\sqrt{2}$): $t = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$

Numerical evaluation ($\sqrt{2} \approx 1.414$):

• Ascent threshold ($t_1$): $t_1 = 2 - \sqrt{2} \approx 0.59\text{ seconds}$
• Descent threshold ($t_2$): $t_2 = 2 + \sqrt{2} \approx 3.41\text{ seconds}$

Step 3: Determine the target interval

For an upward-opening parabola $t^2 - 4t + 2 \le 0$, the inequality holds on the closed interval between its roots: $2 - \sqrt{2} \le t \le 2 + \sqrt{2}$

Step 4: Check apex and boundaries

• At $t = 2 - \sqrt{2} \approx 0.586\text{ s}$: $h(0.586) = 40\text{ m}$ $\checkmark$
• At apex $t = 2\text{ s}$: $h(2) = -5(4) + 20(2) + 30 = 50\text{ m} \ge 40\text{ m}$ $\checkmark$
• At $t = 2 + \sqrt{2} \approx 3.414\text{ s}$: $h(3.414) = 40\text{ m}$ $\checkmark$

$\boxed{\text{Time Interval: } [2 - \sqrt{2}, \; 2 + \sqrt{2}] \text{ seconds} \quad (\approx [0.59\text{ s}, \; 3.41\text{ s}])}$