The distance an object falls is directly proportional to the square of the time it has been falling.?

Question

After 6 seconds it has fallen 1296 ft. how long will it take to fall 2304 ft?

Answer 1

It's possible that there's a mistake in the question as set; if not, the statement of the problem could be improved for clarity:

If this "falling" is not taking place in Earth's own gravity field, it would be customary for a good problem poser to state or imply that fact, for example by saying "On the surface of a newly discovered planet, the distance ... ." (This makes good sense by alerting the student to the fact that the acceleration or 'g' in this problem ISN'T Earth's! "Falling," unqualified, generally implies near Earth's surface, as untold numbers of problems suggest: "A man falls from a cliff 100 feet high ..." Has anyone ever SERIOUSLY asked themselves "On which planet?" --- NO!))

As it happens, with 'g' given by a convenient approximation to Earth's own (surface) 'g' value of 32 ft sec^(-2), the distance fallen after 9 secs (not 6 secs, an inverted '9') would be:

1/2 x 32 x 9^2 ft = 1296 ft.

In any case, the ratio of the two SQUARED t's is the ratio of the two distances, 2304/1296 = 16/9, so the ratio of the two t's themselves is the square root of this, that is: 4/3.

(This latter value, 4/3, is the factor by which the time will need to be increased in the second case, whatever the time is to fall the initial 1296 feet.)

So the answer for how long it takes to fall 2304 ft. WITH THE SAME ACCELERATION is 4/3 x 6 secs = 8 secs in some (non-Earth?) environment where it took 6 secs to fall 1296 ft, or 4/3 x 9secs = 12 secs if the environment was in fact near Earth's own surface with a gravitational acceleration of essentially 32 ft sec^(-2).

Notice that it's simply UNNECESSARY to "SOLVE" for the actual value of the acceleration. A good problem setter will be looking for you to hone in on the fact that all you need to do is to find the square root of the distances ratio in order to find the time enlargement ratio. That really shows that you "GET" the t^2 proportionality, and can use it economically to go directly to the heart of the matter, without excess calculation.

Live long and prosper.

Answer 2

Being an old sky diver for many years I have to disagree with the distance fallen in six seconds.

The formula that we used was Time squared x 16 = distance fallen

Therefore: a six second fall would be: T squared x 16 =

6 x 6 x 16 = 576 feet

to fall 2304 feet would be 2304 divided by 16 = 144 and then find the root of 144 which equals 12

But perhaps different bodies could fall at different speeds but that is the rate that a sky diver falls without the parachute open. This speed increases for about ten seconds and then the parachutist remains at a constant speed. To determine how far a parachutist has fallen he/she uses several methods = altimeters, stop watches and counting mentally. This method has worked for us and is accurate

Answer 3

o.ok. rember Newtons try with the feather and the lead ball? both fell on a similar speed which potential gravitational pull is consistent and consequently the speed of a falling merchandise is consistent. in my opinion it truly is right now touching on to time. with a view to calculate, (2304/1296) * 6 = x = 10.667 seconds.

Answer 4

s=kt^2
1296=k*36
k=1296/36
2304=1296/36*t^2
t^2=2304*36/1296
=64
t=rt64
=8 sec

Answer 5

d=k*t^2
d= 1296 =k * 36
k= 36
2304 = d = 36*t^2
64 = t^2
t = 8 seconds

Answer 6

Yes.

d1 / t1^2 = d2 / t2^2

1296 / 6^2 = 2304 / T^2

Answer 7

at least 12.7 seconds

Answer 8

yes it is, formula is d = 0.5*9.8t^2 meters