When a ball is thrown vertically upward
into the air with a velocity of 79 ft/sec its
height, y(t), in feet after t seconds is given by
y(t) = 79t − 16t
2
.
Find the average velocity of the ball over
the interval from 3 to 3 + h seconds, h 6=/= 0.
1. average vel. = −(17 − 16h) ft/sec
2. average vel. = −(17 + 16h) ft/sec
3. average vel. = −(21 + 16h) ft/sec
4. average vel. = −(21 − 16h) ft/sec
5. average vel. = −(17 − h) ft/sec
6. average vel. = −(21 + h) ft/sec
To find the average velocity of the ball over the interval from 3 to 3 + h seconds, we need to find the displacement of the ball during this time interval and divide it by the time elapsed.
Given the position function y(t) = 79t - 16t^2, we can find the displacement by subtracting the initial position (y(3)) from the final position (y(3 + h)).
Initial position (y(3)):
y(3) = 79(3) - 16(3)^2
= 237 - 16(9)
= 237 - 144
= 93 ft
Final position (y(3 + h)):
y(3 + h) = 79(3 + h) - 16(3 + h)^2
= 237 + 79h - 16(9 + 6h + h^2)
= 237 + 79h - 144 - 96h - 16h^2
= 237 - 144 - 16h^2 + 79h - 96h + 79h
= 93 - 16h^2 - 17h
Displacement (Δy) = y(3 + h) - y(3)
= 93 - 16h^2 - 17h - 93
= -16h^2 - 17h ft
The time elapsed is (3 + h) - 3 = h seconds.
Average velocity = displacement / time elapsed
= (-16h^2 - 17h) / h
= -16h - 17 ft/sec.
Therefore, the correct answer is option 2. average vel. = -(17 + 16h) ft/sec.
You know that the average velocity is
[y(3+h)-y(3)]/h
So, just plug in the numbers and grind out the answers.