Asked by Mae
Find the distance a ball covered in a specific time if it was dropped from the top of a building that 324 meter high. Assume if the gravitational acceleration is 10 meters per second squared. I filled everything except the question marks in the table.
[Seconds] 1 ~ 2 ~ 3
[Distance covered from the beginning of motion] 5 ~ 20 ~ 45
[Distance covered in the last second] 5 ~ ? ~ ?
Please help since I'm home-schooled and my answers are to be input in a computer, but everything except the question marks are correct. Sadly this question has a definite answers that even my teacher can't answer.
[Seconds] 1 ~ 2 ~ 3
[Distance covered from the beginning of motion] 5 ~ 20 ~ 45
[Distance covered in the last second] 5 ~ ? ~ ?
Please help since I'm home-schooled and my answers are to be input in a computer, but everything except the question marks are correct. Sadly this question has a definite answers that even my teacher can't answer.
Answers
Answered by
Reiny
Your quadratic must have been
D = -5t^2 + 324
If we just care about the distance covered
d = 5t^2
so when t = 1, d = 5
when t = 2 , d = 20
when t = 3 , d = 45
You had those correct.
distance covered in the last second:
Since your last second is the third second
take the distance covered after 3 seconds - distance covered after 2 seconds
= 45m - 20 m
= 25 m
D = -5t^2 + 324
If we just care about the distance covered
d = 5t^2
so when t = 1, d = 5
when t = 2 , d = 20
when t = 3 , d = 45
You had those correct.
distance covered in the last second:
Since your last second is the third second
take the distance covered after 3 seconds - distance covered after 2 seconds
= 45m - 20 m
= 25 m
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