Suppose a ball was dropped from 200 ft high. Don’t allow for wind resistance or drag.
How to figure how far it will have fallen and how long it took from any point within that 200 ft. I know 1 g accelerates at about 32 fps. I couldn’t find the formula on line.
An example would be I shoot an arrow at 180 fps, it will travel 60 ft before it hits target. How much will it have dropped?
The general formula is x = x[sub]0[/sub] + v[sub]0[/sub]t + 1/2 at[sup]2[/sup]. If you’re only interested in distance traveled, and if the initial velocity is 0 (as suggested by “dropped”), then it simplifies to Δx = 1/2 at[sup]2[/sup].
For your arrow problem, we must first find the time. Which is really easy, because 60 ft/180 fps is just 1/3 s. And as long as there’s no air resistance (and all of the other standard assumptions), the horizontal and vertical motions are independent.
Distance equals 1/2 the acceleration times the time squared. After one second the ball would have dropped 0.532.2 ft/sec^21 sec^2, so 16 feet. After 3 seconds it will have dropped 0.532.2 ft/sec^23 sec^2, so 144.9 ft. The full 200 foot drop will take a little over 3,5 seconds and reaches 113 feet/sec, neglecting wind resistance.
Thanks, exactly what I needed!
Except those two examples don’t really match up. The first one you’re given distance and need to find the time and in the second one you know the time and need to find the distance.
In the first example you are solving 16.1t[sup]2[/sup] = 200 for t giving t = 3.5245 seconds; in the second example you are solving 16.1(1/3)[sup]2[/sup] = d for d giving d = 1.7889 feet
That is true, but I was really only concerned about the answer as it relates to falling. I used the above example with the arrow because it is an example of something I would actually be looking at.
This makes the answer much easier to calculate - and ensures that it will be wrong by a fair amount.
Googling suggests that a typical arrow has lost something like 10% of its velocity at 60 yds, which means it will have fallen something like 20% further.
I know how to allow for those kind of calculations so wasn’t worried about them.
There are three standard kinematic equations that allow nearly everything you’d want in this type of scenario. Where s is distance, u is initial velocity, v is final velocity, a is acceleration, and t is time you have
v = u + at
s = ut + 1/2at^2
v^2 = u^2 + 2as
Knowing any three of the five variables allows you to calculate the missing two, using one or more of the formulae above (possibly needing some algebraic rearrangement).
I’m surprised that nobody has mentioned that these are Newton’s Equations of motion.
Be careful, because air resistance applies both vertically and horizontally. Even if you use air resistance to get the correct time of flight, you also have to use air resistance to get the correct distance fallen in that time. Worse, air resistance also includes cross terms, meaning that you can’t separate out the vertical and horizontal parts any more: Everyone knows that a dropped bullet and a bullet fired horizontally will both hit the ground at the same time, but that’s not true any more if the bullet has non-negligible air resistance.