Asked by faith
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.
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.
Answers
Answered by
Steve
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
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
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