To determine how far an object travels horizontally, we need to calculate the time of flight (how long the object is in the air) and use the horizontal velocity.

Given:

• Horizontal velocity ($v_x$) = $3.0 \, \text{m/s}$
• Height of the cliff ($h$) = $20 \, \text{m}$
• Acceleration due to gravity ($g$) = $9.8 \, \text{m/s}^2$

Step 1: Calculate the time of flight

The time of flight depends only on the vertical motion. Using the equation of motion:

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

Rearranging to solve for $t$ (time):

$t = \sqrt{\frac{2h}{g}}$

Substitute $h = 20 \, \text{m}$ and $g = 9.8 \, \text{m/s}^2$:

$t = \sqrt{\frac{2(20)}{9.8}} = \sqrt{\frac{40}{9.8}} \approx \sqrt{4.08} \approx 2.02 \, \text{s}$

The object is in the air for approximately $2.02 \, \text{s}$.

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Step 2: Calculate horizontal distance

Horizontal distance is given by:

$d = v_x \cdot t$

Substitute $v_x = 3.0 \, \text{m/s}$ and $t = 2.02 \, \text{s}$:

$d = 3.0 \cdot 2.02 \approx 6.06 \, \text{m}$

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Final Answer:

The object travels approximately 6.06 meters horizontally.

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Follow-up Questions:

1. How would the horizontal distance change if the speed were doubled?
2. What happens to the time of flight if the height of the cliff increases?
3. Can air resistance significantly affect the horizontal distance for this case?
4. How does the acceleration due to gravity influence the result?
5. How would the calculation change if the object were launched at an angle?

Tip: In projectile motion, horizontal and vertical motions are independent, so treat them separately when solving problems.