Answer:
- 2.915 seconds
- 46.648 feet per second
- 1.458 seconds after start
Explanation:
You want to know the air time, the average velocity, and the time to average velocity for a rock dropped from a height of 136 ft. The height is described by s(t) = -16t² +136.
Air time
The rock will hit the ground at the time t that makes s(t) = 0.
s(t) = -16t² +136
0 = -16t² +136
16t² = 136
t² = 136/16
t = √(136/16) ≈ 2.915 . . . . seconds
The rock will hit the ground after 2.915 seconds.
Average velocity
The rock covers the distance 136 feet in 2.915 seconds, so its average velocity is ...
v = (136 ft)/(2.915 s) ≈ 46.648 ft/s
The average velocity is 46.648 ft/s (downward).
Time to average velocity
The velocity is increasing at a constant rate from 0 until the rock hits the ground. Hence the average velocity will be half the final velocity, and will occur after half the time has passed.
The average velocity is reached about 1.458 seconds after the rock is dropped.