Question

For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square of the time. If an object is dropped from the top of a tower 512512 feet high and hits the ground in 88 ​seconds, how far did it fall in the first 55 ​seconds?

Answer 1

Answer:

The distance is 199.98 ft.

Explanation:

Given that,

Distance = 512 feet

Time = 88 sec

The distance the body falls varies directly as the square of the time.

$s\propto t^2$

$s=kt^2$

We need to calculate the value of constant

Put the value into the formula

$512=k\times(88)^2$

$k=\dfrac{512}{(88)^2}$

$k=0.06611$

We need to calculate the distance

Using formula of distance again

$s=kt^2$

Put the value into the formula

$s=0.06611\times(55)^2$

$s=199.98\ ft$

Hence, The distance is 199.98 ft.