To solve this problem, we can break the motion of the boulder into two parts: horizontal and vertical.

Step 1: Vertical Motion

In the vertical direction, the only force acting on the boulder is gravity. We can use the following kinematic equation to determine the time it takes for the boulder to fall from the cliff:

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

Where:

• $y = 47.9 \, \text{m}$ (height of the cliff),
• $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity),
• $t$ is the time in seconds.

Solving for $t$:

$47.9 = \frac{1}{2} \times 9.81 \times t^2$

$t^2 = \frac{47.9 \times 2}{9.81}$

$t^2 = 9.765$

$t = \sqrt{9.765} \approx 3.12 \, \text{seconds}$

Step 2: Horizontal Motion

Now that we know the boulder is in the air for $3.12 \, \text{seconds}$, we can calculate how far it travels horizontally. The horizontal velocity of the boulder is constant at $9.83 \, \text{m/s}$, so we use:

$\Delta x = v_x \cdot t$

Where:

• $v_x = 9.83 \, \text{m/s}$,
• $t = 3.12 \, \text{seconds}$.

$\Delta x = 9.83 \times 3.12 \approx 30.68 \, \text{meters}$

Answer:

The boulder lands 30.68 meters from the base of the cliff.

Would you like more details or have any questions?

Related Questions:

1. How does increasing the height of the cliff affect the distance traveled by the boulder?
2. What would happen if the boulder had an initial vertical velocity as well?
3. How would air resistance alter the result of this problem?
4. How do you calculate the velocity of the boulder when it hits the ground?
5. How would the solution change if the acceleration due to gravity was not constant?

Tip:

Always break down projectile motion problems into horizontal and vertical components to simplify your calculations.