Question

The Empire State Building is about 381m high. If you threw a baseball downward at 25 m/s from the top of the building how fast would it be traveling just before it landed?

Answer 1

Answer:

Explanation:

You can use that $v = v_{o}+gt$ or $v^{2}=v_{o}^{2}+2gh=(25)^{2}+(2)(10)(381)=8245 \rightarrow v \approx 90.8$ m/s.