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 280 feet in the first 5 seconds. How far will it have fallen at the end of 6 seconds? Round your answer to the nearest integer if necessary.

Answer 1

we have that

If the distance varies directly as the square of the time

then the equation is equal to

`d=kt^2`

where k is the constant of proportionality

step 1

Find the value of k

we have

d=280 ft and t=5 sec

substitute

280=k(5^2)

k=280/25

k=11.2

The equation is equal to

d=11.2t^2

step 2

Find the value of d when the value of t is 6 sec

substitute the given value in the equation

d=11.2(6^2)

d=403.2 ft

Round to the nearest integer

d=403 ft