The distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the ground). If an object fel 6 seconds, how far will it have fallen by the end of 8 seconds? (Leave the variation constant in fraction form or round to at least 2 decimal g

Question

Round your final answer to the nearest foot.)

Answer 1

Let's assume that the distance fallen is represented by x, and the time fallen is represented by t.

According to the given information, we can write the equation x = kt^2, where k is the variation constant.

To find the value of k, we need to use the information provided. We know that when t = 6 seconds, x = ? (unknown distance fallen).

Substituting the values into the equation: x = k(6)^2
x = 36k

Now, we can use this equation to find the value of k.

Next, we are asked to find the distance fallen when t = 8 seconds.

Using the same equation: x = kt^2
x = k(8)^2

Substituting the value of k obtained earlier: x = (36k)(8)^2
x = 2304k

Therefore, when t = 8 seconds, the object will have fallen a distance of 2304k.

Since we are not given the value of k, we cannot find the exact distance fallen. We can only state that the object will fall a distance of 2304k.

To convert this to feet (as requested), we would need the value of k.