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.

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

To find the distance covered by the ball in the last second, we can analyze the information given in the table. We know that the ball was dropped from a height of 324 meters, and the gravitational acceleration is 10 meters per second squared.

Let's break down the problem step by step:

1. In the first second, we see that the ball covered a distance of 5 meters. This is the initial velocity of the ball, as it is dropped from rest. We can calculate the final velocity using the equation:

Final velocity = Initial velocity + (Acceleration × Time)

Given that the initial velocity is 0 (as the ball is dropped from rest) and the acceleration is 10 m/s^2, we can calculate the final velocity for the first second:

Final velocity = 0 + (10 × 1) = 10 m/s

2. We can determine the distance covered in the second second using the average velocity:

Average velocity = (Initial velocity + Final velocity) / 2

Since we know the initial velocity is 10 m/s and the final velocity is still unknown, we can rearrange the equation to solve for the final velocity:

Final velocity = 2 × Average velocity - Initial velocity

Plugging in the known values:

Final velocity = 2 × 20 - 10 = 30 m/s

Now that we have the final velocity for the second second, we can calculate the distance covered in that second using the same equation as before:

Distance covered = Initial velocity × Time + (1/2) × Acceleration × Time^2

Distance covered = 10 × 1 + (1/2) × 10 × 1^2 = 10 + 5 = 15 meters

3. Finally, for the third second, we need to calculate the final velocity, then the distance covered:

Final velocity = 2 × Average velocity - Initial velocity

Given that the average velocity is 30 m/s and the initial velocity is 10 m/s:

Final velocity = 2 × 30 - 10 = 50 m/s

Distance covered = Initial velocity × Time + (1/2) × Acceleration × Time^2

Distance covered = 10 × 2 + (1/2) × 10 × 2^2 = 20 + 20 = 40 meters

Let's fill in the corresponding distances in the table:

[Seconds] 1 ~ 2 ~ 3
[Distance covered from the beginning of motion] 5 ~ 20 ~ 45
[Distance covered in the last second] 5 ~ 15 ~ 40

So, the distance covered in the last second is 15 meters for the second second and 40 meters for the third second.