Question

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?

Answer 1

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 .

Answer 2

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.