To determine how far from the base of the cliff the stone hits the ground, we need to analyze both the horizontal and vertical components of the stone’s motion. This is a classic projectile motion problem where the stone is thrown horizontally, so the initial vertical velocity is zero.

Step-by-Step Solution:

1. Vertical Motion (free-fall):

   • The vertical distance ($h$) the stone falls is 78.4 m.

   • The initial vertical velocity ($v_{y0}$) is 0 (since it’s thrown horizontally).

   • The acceleration due to gravity ($g$) is $9.8 \, \text{m/s}^2$.

   • Using the formula for distance in free-fall:
     $h = \frac{1}{2} g t^2$ we can solve for the time ($t$) it takes to hit the ground.

     $78.4 = \frac{1}{2} \times 9.8 \times t^2$ $t^2 = \frac{78.4}{4.9} = 16$ $t = \sqrt{16} = 4 \, \text{seconds}$

2. Horizontal Motion:

   • The horizontal velocity ($v_x$) remains constant at 8 m/s (since there’s no horizontal acceleration).
   • The horizontal distance ($d$) can be found using the formula: $d = v_x \times t$ $d = 8 \times 4 = 32 \, \text{meters}$

Answer:

The stone will hit the ground 32 meters from the base of the cliff.

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Would you like additional details, or have any questions about this solution?

Here are 5 related questions to deepen understanding:

1. What would be the impact point if the stone was thrown at a different speed, say 10 m/s?
2. How would the impact distance change if the cliff height increased to 100 meters?
3. What would the result be if the stone was thrown at an angle instead of horizontally?
4. How does air resistance affect the stone’s impact point in real scenarios?
5. Can we calculate the stone's final velocity upon hitting the ground?

Tip: In horizontal projectile motion, remember that vertical and horizontal motions are independent, with gravity only affecting the vertical component.