Let's solve the problem step by step.

Given:

• Initial horizontal speed of the stone, $v_x = 5.0 \, \text{m/s}$
• Height of the cliff, $h = 78.4 \, \text{m}$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

a) Time to reach the bottom of the cliff

Since the stone is thrown horizontally, its initial vertical velocity is 0. The time it takes to reach the ground is determined solely by the vertical motion. We can use the following equation for vertical displacement:

$h = \frac{1}{2} g t^2$

Solving for time $t$:

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 78.4 \, \text{m}}{9.8 \, \text{m/s}^2}}$

$t \approx \sqrt{16} = 4.0 \, \text{seconds}$

So, it takes $t = 4.0 \, \text{seconds}$ for the stone to reach the bottom of the cliff.

b) Horizontal distance from the base of the cliff

The horizontal distance $d_x$ can be calculated using the horizontal velocity $v_x$ and the time $t$:

$d_x = v_x \times t = 5.0 \, \text{m/s} \times 4.0 \, \text{seconds} = 20.0 \, \text{m}$

The stone strikes the ground 20.0 meters away from the base of the cliff.

c) Final speed of the stone as it hits the ground

The final speed of the stone has both horizontal and vertical components. The horizontal speed $v_x$ remains constant at 5.0 m/s. The vertical speed $v_y$ at the moment of impact can be found using:

$v_y = g \times t = 9.8 \, \text{m/s}^2 \times 4.0 \, \text{seconds} = 39.2 \, \text{m/s}$

The final speed $v_f$ is the magnitude of the vector sum of $v_x$ and $v_y$:

$v_f = \sqrt{v_x^2 + v_y^2} = \sqrt{(5.0 \, \text{m/s})^2 + (39.2 \, \text{m/s})^2}$

$v_f \approx \sqrt{25 + 1536.64} = \sqrt{1561.64} \approx 39.5 \, \text{m/s}$

Thus, the final speed of the stone as it hits the ground is approximately 39.5 m/s.

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Do you want more details on any of these steps, or do you have any questions?

Here are some related questions:

1. How would the time to reach the ground change if the cliff were twice as high?
2. What would happen to the horizontal distance if the initial speed were doubled?
3. How would air resistance affect the stone's trajectory?
4. How can we calculate the angle of impact of the stone?
5. What would be the final speed if the stone were thrown at an angle instead of horizontally?

Tip: In projectile motion problems, the horizontal and vertical motions are independent of each other, except for being connected by the time of flight.