What’s the unit rate of 4,500 kilobytes in 6 minutes
1 answer:
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Answer:
a) The height of the small object 3 seconds after being launched is 2304 feet.
b) The small object ascends 128 feet between 5 seconds and 7 seconds.
c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.
d) The object will take 21 seconds to hit the ground.
Step-by-step explanation:
The correct formula for the height of the small object is:
(1)
Where:
- Height above the ground, measured in feet.
- Time, measured in seconds.
a) The height of the small object at given time is found by evaluating the function:
The height of the small object 3 seconds after being launched is 2304 feet.
b) First, we evaluate the function at and
:
We notice that the small object ascends in the given interval.
The small object ascends 128 feet between 5 seconds and 7 seconds.
c) If we know that , then (1) is reduced into this second-order polynomial:
(2)
All roots of the resulting equation come from the Quadratic Formula:
and
The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.
d) If we know that , then (1) is reduced into this second-order polynomial:
(3)
All roots of the resulting equation come from the Quadratic Formula:
and
Just the first root offers a solution that is physically reasonable.
The object will take 21 seconds to hit the ground.