Posted by **faith** on Thursday, December 15, 2011 at 9:46pm.

A person standing on the roof of a building throws a ball directly upward. The ball misses the rooftop on its

way down and eventually strikes the ground. The function s(t) = −16t2 + 64t + 80 describes the ball’s height

above the ground, s(t) , in feet, t , seconds after it was thrown.

a. Find the ball’s average velocity between the time it was thrown and 2 seconds later.

b. Find the ball’s average velocity between 2 and 4 seconds after it was thrown.

- math -
**Steve**, Friday, December 16, 2011 at 11:39am
Since v(t) = 64 - 32t

v(2) = 0

So, the velocity decreases from 64 to 0 in 2 seconds. Average v = 32

Similarly, once the ball starts falling, it regains its velocity of -64 after 2 seconds. So, on the way down, avg v = -32

## Answer This Question

## Related Questions

- physics, please help me, this question is hard - John Stands at the edge of a ...
- Physics - Standing on the roof of a 62 m tall building, you throw a ball ...
- Physics - Two students are on a balcony 19.6 m above the street. One student ...
- algebra - A ball is thrown straight up from a rooftop 9696 feet high. The ...
- physics - two students are on a balcony 20.6 m above the street. One student ...
- physics - two students are on a balcony 20.6 m above the street. One student ...
- physics - Two students are on a balcony 21.4 m above the street. One student ...
- physics - Two students are on a balcony 18.3 m above the street. One student ...
- physics - Two students are on a balcony 18.3 m above the street. One student ...
- physics - A person standing on top of a building throws a ball with a horizontal...

More Related Questions