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
Round your final answer to the nearest foot.)
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.