High School

We appreciate your visit to If an object is dropped from a 157 foot high building its position is given by tex s t 16t 2 157 tex When will. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

If an object is dropped from a 157-foot high building, its position is given by [tex]s(t) = -16t^2 + 157[/tex].

When will it hit the ground, and what is the velocity upon impact?

Answer :

Final answer:

The object will hit the ground in approximately 3.94 seconds and the velocity upon impact will be -126.08 feet per second.

Explanation:

If an object is dropped from a 157 foot high building, its position is given by the equation s(t) = -16t² + 157.

To find when it will hit the ground, we need to find the time when the position is equal to zero.

We can solve the equation -16t² + 157 = 0 using the quadratic formula.

The positive root is the one we are interested in, which is t = 3.94 seconds.

To find the velocity upon impact, we can differentiate the position function with respect to time.

The derivative of s(t) = -16t² + 157 is v(t) = -32t.

Plugging in the time

t = 3.94 seconds,

we get v(3.94) = -126.08 feet per second.

Learn more about object's position, time of impact, velocity upon impact here:

https://brainly.com/question/34367613

#SPJ4

Thanks for taking the time to read If an object is dropped from a 157 foot high building its position is given by tex s t 16t 2 157 tex When will. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada