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A marble is released from rest at the top of a high cliff. It falls 4.00 m in the first 1 s of its motion. Through what additional distance does it fall in the next 1 s?

A. 4.00 m
B. 8.00 m
C. 12.0 m
D. 16.0 m
E. 20.0 m

Answer :

Final Answer:

The marble falls an additional distance of (c) 12.0 m in the next 1 second.

Explanation:

In the first second of motion, the marble falls 4.00 meters. This is because the distance fallen under the influence of gravity is calculated using the formula:

[tex]\[d = \frac{1}{2}gt^2\][/tex]

where [tex]\(d\)[/tex] is the distance fallen, [tex]\(g\)[/tex] is the acceleration due to gravity (approximately [tex]\(9.81 \, \text{m/s}^2\)), and \(t\)[/tex] is the time in seconds.

Given that [tex]\(d = 4.00 \, \text{m}\) and \(t = 1 \, \text{s}\)[/tex], we can solve for [tex]\(g\)[/tex]:

[tex]\[4.00 = \frac{1}{2} \times g \times (1)^2\][/tex]

[tex]\[g = 8.00 \, \text{m/s}^2\][/tex]

Now, for the next second of motion [tex](\(t = 2 \, \text{s}\))[/tex], we can use the same formula to calculate the additional distance fallen:

[tex]\[d = \frac{1}{2} \times g \times (2)^2\][/tex]

[tex]\[d = 12.0 \, \text{m}\][/tex]

Thus, the marble falls an additional 12.0 meters in the next 1 second.

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