Question

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 fell 76 ft in 4 seconds, how far will it have fallen by the end of 7 seconds? round your answer to the nearest integer if necessary.

Explanation:

Step1: Set up the proportion equation

Let $d$ be the distance and $t$ be the time. Since $d$ is directly proportional to $t^{2}$, we have $d = kt^{2}$. When $d = 76$ and $t = 4$, we can find $k$. Substitute into the equation: $76=k\times4^{2}$, so $76 = 16k$. Then $k=\frac{76}{16}=\frac{19}{4}$.

Step2: Find the distance at $t = 7$

Now that $k=\frac{19}{4}$, when $t = 7$, we use the equation $d=kt^{2}$. Substitute $k=\frac{19}{4}$ and $t = 7$ into it: $d=\frac{19}{4}\times7^{2}=\frac{19}{4}\times49=\frac{931}{4}=232.75$. Rounding to the nearest integer, $d = 233$.

Answer:

233