Question

The distance that an object falls from rest, when air resistance is negligible, varies directly as the square of the time that it falls (before it hits the ground). A stone dropped from rest travels 193193 feet in the first 44 seconds. How far will it have fallen at the end of 55 seconds? (Round off your answer to the nearest integer.)

Answer 1

Answer:

$\boxed{301865\:feet}$

Step-by-step explanation:

Let the distance be $d$ and the time be $t$.

It was given that the distance, $d$, varies directly as the square of the time $t$, we write the proportional relation,

$d\propto t^2$.

$\Rightarrow d=k t^2$, where $k$ is the constant of variation.

It was also given that, $d=193,193$, when $t=44$.

$\Rightarrow 193193=k (44^2)$

$\Rightarrow 193193=1936k$

$\Rightarrow \frac{193193}{1936}=k$

$\Rightarrow k=99.79$

Now the equation of variation becomes;

$d=99.79t^2$

When, $t=55s$, $d=99.79(55^2)$

$\Rightarrow d=99.79(3025)$

$d=301864.75$

The object will travel 301865 feet in 55 seconds.