The distance a body falls from rest varies directly as the square of the time the body is falling. If a body falls 81 ft. in 3 seconds, how far will it fall in 7 seconds?

Let the distance = D

Let the time taken = T

From the question, we are told that distance is directly proportional to the square of the time the body takes to fall,

that is, D = K * [T^2]

K is a constant and we have to find its value;

Using the value given in the question;

D = 81 feet

T = 3 seconds

From our original equation,

K = D / [T^2] = 81/3^2 = 81/9 = 9.

Therefore, K = 9

To find the distance through which the body fall in 7 seconds, we have,

K = 9

T = 7 seconds

D = K * [T^2]

D = 9 * [7*2] = 9 * 49 = 441.

Therefore, the body will fall through a distance of 441 feet in 7 seconds.

You are saying that X = k*t^2 where k is a constant

Plugging in the 64 ft and 2 seconds k = 16 so,

X = 16*49 = 784 ft.