To find the time it takes for an object thrown straight down to reach the ground, we can use the following kinematic equation:

$s = v_0 t + \frac{1}{2} a t^2$

where:

• $s$ is the distance (5.48 meters),
• $v_0$ is the initial velocity (8.17 m/s),
• $a$ is the acceleration due to gravity (approximately 9.81 m/s²),
• $t$ is the time in seconds.

Rewriting the equation:

$5.48 = 8.17 t + \frac{1}{2} \cdot 9.81 \cdot t^2$

This simplifies to:

$5.48 = 8.17 t + 4.905 t^2$

This is a quadratic equation in the form:

$4.905 t^2 + 8.17 t - 5.48 = 0$

To solve for $t$, we use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 4.905$, $b = 8.17$, and $c = -5.48$.

Plugging in the values:

$t = \frac{-8.17 \pm \sqrt{(8.17)^2 - 4 \cdot 4.905 \cdot (-5.48)}}{2 \cdot 4.905}$

$t = \frac{-8.17 \pm \sqrt{66.7489 + 107.3328}}{9.81}$

$t = \frac{-8.17 \pm \sqrt{174.0817}}{9.81}$

$t = \frac{-8.17 \pm 13.1925}{9.81}$

This gives us two solutions:

$t_1 = \frac{-8.17 + 13.1925}{9.81} \approx \frac{5.0225}{9.81} \approx 0.512 \, \text{seconds}$

$t_2 = \frac{-8.17 - 13.1925}{9.81} \approx \frac{-21.3625}{9.81} \approx -2.178 \, \text{seconds}$

Since time cannot be negative, we discard $t_2$.

Therefore, the time it takes to reach the ground is approximately:

$t \approx 0.512 \, \text{seconds}$

Would you like further details or have any questions?

Here are 5 numbered questions you might ask:

1. How do you derive the kinematic equations used in physics?
2. Can you explain more about solving quadratic equations?
3. What assumptions are made in this type of problem?
4. How does air resistance affect the time to reach the ground?
5. Can you show the step-by-step process of using the quadratic formula?

Tip: Always check the units of your given values to ensure they are consistent before performing calculations.