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Wile E. Coyote drops a bowling ball off a cliff that is 132 meters high. How fast is it going at the end of 3.0 seconds?

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Final answer:

The bowling ball is going at 29.4 m/s after falling for 3.0 seconds, calculated using the formula v = gt, where g is 9.8 m/s².

Step-by-step explanation:

To determine how fast the bowling ball is going at the end of 3.0 seconds after being dropped off a cliff, we use the equation of motion under constant acceleration due to gravity. We will use the formula v = gt, where v is the final velocity, g is the acceleration due to gravity, and t is the time the object has been falling.

Assuming the standard acceleration due to gravity as 9.8 m/s², the calculation becomes v = 9.8 m/s² * 3.0 s. Therefore, the final velocity v would be 29.4 m/s.

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