A ball is dropped from the top of a building. The balls fall 3 times farther during its last second of freefall than it did during its third second of freefall. The ball fell 4.9 meters during its first second, how tall is the building and how much time is the ball in the air?

Answer d = 313.6 m

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313.6 To find the height of the building and the time the ball is in the air, we can use the equations of motion and analyze the problem step by step.

1. Let's first look at the distance the ball falls during its second and third seconds of freefall.

During the second second, the ball is under constant acceleration, so we can use the equation:
d2 = (1/2) * g * t2^2

During the third second, the distance is three times less than the distance during the last second of freefall. So, by using the same equation:
d3 = (1/2) * g * t3^2
d3 = (1/2) * g * (t3 - 1)^2
d3 = (1/2) * g * (t3^2 - 2t3 + 1)
3d3 = (1/2) * g * (t3^2 - 2t3 + 1)

Now, we have a relationship between the distance during the third second (3d3) and the distance during the last second of freefall (d2).

2. We know that the distance during the first second of freefall is 4.9 meters. So, we can write the equation for the distance during the third second in terms of t3:
3(4.9) = (1/2) * g * (t3^2 - 2t3 + 1)
14.7 = (1/2) * g * (t3^2 - 2t3 + 1)

3. We can simplify this equation:
14.7 = (1/2) * g * t3^2 - g * t3 + (1/2) * g
14.7 = (1/2) * g * t3^2 - g * t3 + 4.9

4. Now, we can rearrange the equation to solve for g * t3^2 - 2g * t3:
g * t3^2 - 2g * t3 = 14.7 - 4.9
g * t3^2 - 2g * t3 = 9.8

5. We can divide both sides of the equation by g to get:
t3^2 - 2t3 = 9.8/g

6. Since we know that g = 9.8 m/s^2, we can substitute it into the equation:
t3^2 - 2t3 = 1

7. Now, we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Factoring in this case gives us:
(t3 - 1) * (t3 - 1) = 0

8. Solving for t3:
t3 - 1 = 0
t3 = 1

9. Now that we have found t3, we can substitute it back into the equation to find the distance during the last second of freefall:
d2 = (1/2) * g * t3^2
d2 = (1/2) * 9.8 * 1^2
d2 = 4.9 meters

10. We have found that the distance during the last second of freefall is also 4.9 meters, which means that the ball falls 3 times farther during its last second of freefall than during its third second.

11. Since the height of the building is the total distance the ball travels during its fall, we can calculate it:
d = d1 + d2 + d3 + ...

Using the formula for the distance traveled during each second of freefall, we find that:
d = (1/2) * g * (1^2) + (1/2) * g * (2^2) + (1/2) * g * (3^2) + ...
d = (1/2) * 9.8 * (1^2 + 2^2 + 3^2 + ...)
d = (1/2) * 9.8 * (1 + 4 + 9 + ...)

12. This is the sum of the squares of consecutive positive integers, which is a known sequence. The formula for the sum of squares of consecutive positive integers is:
1^2 + 2^2 + 3^2 + ... + n^2 = (n * (n + 1) * (2n + 1)) / 6

13. Substituting n = 3 into the formula, we get:
1^2 + 2^2 + 3^2 = (3 * (3 + 1) * (2 * 3 + 1)) / 6
1 + 4 + 9 = (3 * 4 * 7) / 6
14 = 42 / 6
14 = 7

14. Substituting this back into the equation for the distance, we find:
d = (1/2) * 9.8 * 7
d = 34.3 meters

Therefore, the height of the building is approximately 34.3 meters, and the ball is in the air for approximately 3 seconds.