An object is released from rest at t = 0 at the top of a cliff.How far does it fall during the second second of its fall (i.e., from t = 1 s to t = 2 s)?Neglect friction and express your answer in meters.
Question
An object is released from rest at t = 0 at the top of a cliff.How far does it fall during the second second of its fall (i.e., from t = 1 s to t = 2 s)?Neglect friction and express your answer in meters.
Solution
To solve this problem, we need to use the equations of motion. The distance an object falls under gravity is given by the equation d = 0.5gt^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth) and t is the time.
First, we calculate how far the object falls in the first 2 seconds: d(2) = 0.5 * 9.8 * 2^2 = 19.6 m
Then, we calculate how far the object falls in the first second: d(1) = 0.5 * 9.8 * 1^2 = 4.9 m
The distance the object falls in the second second is the difference between these two distances: d(second second) = d(2) - d(1) = 19.6 m - 4.9 m = 14.7 m
So, the object falls 14.7 meters during the second second of its fall.
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