To solve this problem, we can use the following kinematic equation for vertical motion:

$s(t) = s_0 + v_0 t - \frac{1}{2} g t^2$

Where:

• $s(t)$ is the height of the object at time $t$,
• $s_0 = 256 \, \text{ft}$ is the initial height (the height of the cliff),
• $v_0 = 64 \, \text{ft/sec}$ is the initial velocity (upward),
• $g = 32 \, \text{ft/sec}^2$ is the acceleration due to gravity (in ft/sec²),
• $t$ is the time in seconds.

The object will hit the ground when $s(t) = 0$ (the height is zero).

Thus, we set up the equation:

$0 = 256 + 64 t - 16 t^2$

Now, we solve this quadratic equation:

$16t^2 - 64t - 256 = 0$

Step 1: Simplify the equation

Divide through by 16:

$t^2 - 4t - 16 = 0$

Step 2: Solve using the quadratic formula

The quadratic formula is:

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

For the equation $t^2 - 4t - 16 = 0$, we have:

• $a = 1$,
• $b = -4$,
• $c = -16$.

Substitute these values into the quadratic formula:

$t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-16)}}{2(1)}$

$t = \frac{4 \pm \sqrt{16 + 64}}{2}$

$t = \frac{4 \pm \sqrt{80}}{2}$

$t = \frac{4 \pm 4\sqrt{5}}{2}$

$t = 2 \pm 2\sqrt{5}$

Step 3: Calculate the two solutions

We get two possible values for $t$:

$t = 2 + 2\sqrt{5} \quad \text{or} \quad t = 2 - 2\sqrt{5}$

Since time cannot be negative, we discard the negative solution.

Now, $t = 2 + 2\sqrt{5} \approx 6.472$ seconds.

Therefore, the object will hit the ground after approximately 6.47 seconds.

Would you like more details or have any questions?

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Here are 5 relative questions:

1. How would the time change if the initial velocity was different?
2. What if the height of the cliff was doubled? How long would it take then?
3. How would the equation change if the object was thrown downward?
4. How high would the object go before starting to fall?
5. What is the velocity of the object when it hits the ground?

Tip: The quadratic formula is a powerful tool for solving any second-degree equation, but you can also use factoring when it's simpler.