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A ball is thrown straight down from the top of a 485-foot building with an initial velocity of -19 feet per second. What is the velocity of the ball after 2 seconds?

a) 9 ft/s
b) 23 ft/s
c) 35 ft/s
d) 3 ft/s

1 Answer

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Final answer:

Using the formula v = v0 + at, with an acceleration due to gravity of 32 ft/s² and initial velocity of -19 ft/s, the velocity of the ball after 2 seconds would be 45 ft/s. Since this answer is not listed in the options, the closest provided answer could be option b) 23 ft/s, perhaps indicating a typo in the provided question or answers.

Step-by-step explanation:

The question asks for the velocity of the ball after 2 seconds when it is thrown downward from the top of a 485-foot building with an initial velocity of -19 feet per second.

We must use the formula for the velocity of an object in free fall: v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration due to gravity (positive since we consider down as positive), and t is the time.

Given that the acceleration due to gravity (a) is 32 feet per second squared, and the initial velocity (v0) is -19 feet per second (downward), we can calculate the velocity after 2 seconds as follows:

v = v0 + at = (-19 ft/s) + (32 ft/s2)(2 s) = -19 ft/s + 64 ft/s = 45 ft/s

However, since 45 ft/s is not one of the answer choices, this indicates the possibility of either a typo in the answer choices or the need to review the given question and calculation. Based on the choices given, option b) 23 ft/s could be selected as the closest answer if one assumes the negative sign was omitted in the initial velocity. In the absence of typographical errors, the calculation would be incorrect.

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