Cannon Ball Conundrum
A cannon ball is shot out of a cannon and travel 4 miles before it hits the ground, at the same time a cannon ball is dropped out of the sky 4 miles off the ground
The question is Which Cannon ball will hit the ground first, Explain
The question is Which Cannon ball will hit the ground first, Explain
posted by Eric at
5:38 PM
38 Comments:
I beleive the key here is something called terminal velocity. assuming the cannon would have to fire at a 45 degree angle to get the required 4 miles. Both the falling ball and the projectile would have the same terminal velocity once the energy that was created by the cannon was expended. which means both balls will hit the ground at the same time...i think?
This can't be answered without knowing the angle at which the ball left the cannon, or maybe the muzzle velocity. E.g. the angle could be almost vertical and rise, say 10 miles before dropping and landing 4 miles away. in this case the falling ball would win easily. On the other hand, if the cannon barely skimmed the ground before landing at the 4 mile mark, then would this be quicker? This needs some applied maths to solve.
Chris
The only time the question has a deterministic answer with the given data is when this is in vacuum and the canon is fired at 45 degrees. Otherwise you need to know drag coefficients, muzzle velocities, etc.
In vacuum at 45 degrees it takes exactly the same amount of time:
Free-fall from height h takes sqrt(2h/g) time.
At 45 deg, there are two equal velocity components, horizontal and vertical. Each is V/sqrt(2) in magnitude, if muzzle velocity is V.
Vertical component:
It takes t1 = V/g*sqrt(2) to get to the highest point and the same to get back down to 0 elevation for total time:
t = V*sqrt(2)/g
In this time horizontal travel is
h = V*t/sqrt(2), or
V = h*sqrt(2)/t
put this into the vertical component expression for t and simplify:
t^2 = 2h/g,
which is the same t as for the free-fall case.
Assume there is no friction.
let:
Vo be the muzzle velocity
h be the height of the falling ball
d be the distance of the shot ball
Θ be the angle of the shot above the horizontal
The time it takes for the free-falling ball to hit the ground is given by the formula h=1/2*gt^2.
Solving for t you get:
t=sqrt(2h/g)
For the shot ball, the initial horizontal and vertical component of the ball is Vo*cosΘ and Vo*sinΘ respectively. The final horizontal and vertical component of the ball is Vo*cosΘ and -Vo*sinΘ respectively.
The amount of time that the shot cannon ball is airborne is given by the formula -Vo*sinΘ=Vo*sinΘ-gt; rearranging gives us Vo=gt/2sinΘ. Substituting t for sqrt(2h/g) you get:
Vo= g*sqrt(2h/g)
---------------
2sinΘ
The distance traveled by the shot ball is given by the formula d=Vo*cosΘ*t. Rearranging we get Vo=d/(cosΘ*t). Since it is given h=4 and d=4, we can conclude h=d. Substituting for t and d we get the formula:
Vo = h
-----------------
cosΘ * sqrt(2h/g)
Since these two formula are equal to Vo, therefore
h g*sqrt(2h/g)
----------------------- = -------------
cosΘ * sqrt(2h/g) 2sinΘ
Simplifying this equation, we are left with:
sinΘ=cosΘ
Θ=45°
Therefore if the shot cannonball must land exactly 4 miles from where it was shot, if the angle is at 45° both cannonballs will land at the same time. An angle greater than 45°, the free-falling cannonball will land first. An angle less than 45°, the shot cannonball will land first.
an edit to my above post because it isn't that clear
In conclusion, assuming the shot cannonball must land exactly 4 miles from where it was shot, if shot with an angle of 45° above horizontal, both cannonballs will land at the same time. If shot at an angle greater than 45°, the free-falling cannonball will land first. If shot at an angle less than 45°, the shot cannonball will land first.
Put all your theories together and you come up with 3 "IF" statements:
(B1=ball shot from cannon, B2=ball dropped, ^ =angle)
IF ^ B1 shot at < 45, B1 lands first
IF ^ B1 shot at = 45, B1 and B2 land at the same time
IF ^ B1 shot at > 45, B2 lands first
OMG, um I am almost sure the one in the sky
Forget ifs whys and wherefore, the dropped ball can only go at terminal velocity. To send a cannon ball 4 miles the initial muzzle velocity is far far in excess of terminal velocity and the dropped ball will just never catch up. it would take the dropped ball several minutes to land, Cheers Paul
I estimate the cannon ball will reach
terminal velocity at less than 200 mph,
so if it averages 200 mph, the fall
will take 4 /200 = .02 hours or about
72 seconds. Air friction determines the
terminal which is about 135 mph for a human
and a cannon ball would be faster, so
I used 200 mph. If it averaged 400, it would
be 36 seconds, but I think 200 would be closer.
The distance a cannon ball shot from a
cannon depends on many variables like the
angle, initial velocity, etc. The problem
says it travelled 4 miles, so let's estimate
it travels 1/2 up, 3 miles down range and falls
another 1/2 mile.
T1=SIND(Ang)*V0/G ;time to reach max height
Ymax=(sind(Ang))^2*V0^2/(2*G) ;max height
T2=2*Sind(Ang)*V0/G ;time to impact @ same Elev
Xrange=V0^2*sind(2*Ang)/G
MileX=Xrange/5280
MileY=Ymax/5280
To reach 1/2 mile in elevation, 3 miles range, would take 32.2 degrees and an initial velocity of 743 feet per sec.
and would take 25 seconds.
The cannon ball shot would hit the ground first and take about 26 seconds, the falling ball would take 72 seconds.
Even if it could fall at 400 mph, it would still take longer
than the shot one.
hjg
As a post script to my earlier answer a 105mm abbot would toss a shell with a muzzle velocity of 300 meters per second this would take about 5.8 seconds to land 4 miles away. You could have had coffee and a cigarette before the dropped ball had landed. Cheers Paul
surely it would be the one from the sky. wouldn't it???
As I noted, and Sam explained in excellent detail earlier, this is fairly straightforward without air resistance.
With air resistance things get a bit more complex. I recently wrote a simple simulator for this, so I did some experiments with it and came up with some interesting observations, that I wanted to share
I assume the following for specificity; the results do not vary hugely with changes in these assumptions:
- the cannonball is made of cast iron, density of 7200 kg/m^3 (approximate; should be between 7000 and 7600 in reality, depending on the actual composition)
- the cannonball is around 32 lb, a large naval weapon size, but on the smaller side for stationary cannon. For this weight, it is 8 cm radius (just over 3 in).
- the cannonball is fairly smooth, having a drag factor of 0.1.
- the cannon is fired at the optimal angle for distance
Results:
- terminal velocity is around 350 m/s (800 mph).
- dropping the ball from 4 miles results in the ball hitting the ground at 39.4 sec with velocity of 281 m/s, short of terminal.
- firing the ball for a 4-mile distance has to occur at angle of 41 degrees with muzzle velocity of 314 m/s (10% or so short of terminal). The highest elevation is just over a mile (1675.6 m) at 17.5 sec, distance of 3490 m (2.17 mi), velocity 172 m/s. Landing is at 36.87 sec, velocity 210 m/s, angle -51.5 deg.
Observations:
- times are similar, consistent with fairly low air resistance of the ball.
- highest flight point comes a bit before the mid-point in time; more than half of the distance is covered at this point: the air resistance slows the ball down significantly.
- elevation angle for maximum distance is under 45%. This makes sense, since gravity helps against air resistance in the vertical direction, but not horizontal, so more energy needs to be directed horizontally to overcome air resistance.
- the ball is at its lowest speed at the highest point: gravity speeds up the vertical component faster than the air slows down the horizontal component.
If anyone is interested, I can post or e-mail the detailed sim results or the code (it's in perl, about 150 lines of code).
You forgot one thing!!!!
It has nothing to do with the elevation it has to do with how much powder, the more powder the less elevation the faster the projectile. your treating it like a bowling ball with just enough velocity to get there. you cant catch real cannon balls so it has more velocity than just enough to get there!!! Cheers Paul
they hit the ground at the same time! basic physics tells you that. its like the monkey in the tree and the hunter if the the hunter aims right at the monkey and fires at the exact same time the monkey falls he will always hit the monkey!
Basic physics, Bah, If you fire the ball at the speed of sound, twice the speed of sound, three times or even the speed of light they would hit at the same time. Bah humbug!!
My guess, as one who has had to explain such thought problems in technical job interviews, is it's fair to say *all things being equal* which hits first.
If that assumption is made, then the ball shot from the cannon travels farther than the one dropped from the sky. Therefore, the one from the sky would cover the 4 miles faster.
All the complicated exceptions, assumptions, calculations, etc., aside, am I missing something?
Guess what all you who did complicated equations are WRONG! they would hit at the same time
Since it's your post Eric, would you mind telling us how you figure they'll both land at the same time?
The second anon put it quite succinctly, "the angle could be almost vertical and rise, say 10 miles before dropping and landing 4 miles away. in this case the falling ball would win easily"
they would hit the ground at the same time cuz its still being throwen into the sky and wat the hell is up with people making equations
they will land at the same time because if they are shot at the same time and both ate traveling the same distance. All objects fall at 9.8 meters per second. So they hot at the same time
last anon and cj:
the cannon balls are being shot (dropped) in different directions, one with a parabolic path and one with a straight path. The dropped ball has an acceleration of 9.8m/s, but you don't know the speed of the shot ball because the powder can impart any speed and still have the ball land 4 miles away (just adjust the angle).
Therefore, you assumption that they travel the same distance and/or at the same speed is not specified by the problem and you can't assume it's the same (leading, in turn, to all the math).
Another version of this puzzle specifies that the balls start at the same height and that one is shot horizontally. This is probably what was meant by the question, but was not specified, so cannot be assumed. The most famous version of the question uses a monkey, as mentioned in an earlier post.
i beleive that the ball that is dropped 4 miles well hit the ground frist. only because of the affect of gravity which is that gravity well pull on an object increasing its speed until the object makes comtact with the ground or other.
Ok so:
When the ball that is shot out of a cannon has the advantage of have a large starting velocity while the one dropped out of the sky has the advantage of having a constant speed up, so when you figure it they both have different advantages to help them move along and at the end they will both even out
So, Eric, you're saying that regardless of the angle and initial velocity of cannon-fired ball, the time it takes to hit the ground remains equal to a ball dropped with a constant acceleration of 9.8m/s/s?
So say the cannon fired straight up (almost), and shot with enough powder to shoot the ball 10 miles into the air. You still think it'll "catch up" with ball dropped from a mere 4 miles up? Keep in mind that the shot cannon ball has a significant upward velocity when it passes the 4 miles mark on its way up.
Ok... I think most of your points are valid, just one more bit of information needs to be added to the question: Why would someone in their right mind drop a cannon ball out of the sky??
Suppose the cannon was in a helicopter hovering 4 miles above the ground with the cannon aimed straight at the ground. If the cannon were to fire one ball at the same time as the other one was dropped, the ball from the cannon would reach the ground before the dropped one.
They will both hit the ground at the same time as the pull of gravity is the same for both balls. The ball that is shot out of the cannon travels 4 mile so the adjustments are made to the angle of the cannon and the velocity of the ball so 4 miles is the end result of travel. During the 4 mile flight, gravity is still acting on it the exact same way as the freefall ball.
the one that was dropped will hit the ground first because it will pick up speed as it goes down
I think that both cannonballs will hit the ground at the same time because when the one shot hits it's top point the other one is dropped and science says that even if two objects weigh differently they will hit ground at the same time. So the cannonballs are at the same point dropping and hit the ground at the same tome
oops i meant time not tome =)
nerds blah blah blah
They will hit at the same tine
We need to know the speed imparted to the cannon in vertical direction,ie vsinΘ component to determine the actual time difference
For the cannon dropped its zero but for the one shot its not known
We need to know the speed imparted to the cannon in vertical direction,ie vsinΘ component to determine the actual time difference
For the cannon dropped its zero but for the one shot its not known
maybe you should treat it like a high school physics problem. ignore air resistance and friction for a sec. one will be traveling at 9.8 m/s^2 strait down and the other will be travelling out of a cannon according to the range equation. so it lands on the ground going at the exact same speed it came out of the cannon at. But all things considered the answer is not the same time. Eric if you are going to propose a question make sure you A) do it clearly and B) give enough information.
A problem i heard like this one is that you have a gun and you shoot a bullet horizontally to the ground, and at the EXACT same time, you drop a bullet from the same hight as the gun. assuming a flat earth and all the other wonderful things needed in a physics class, which one will hit the ground first??
it would be the one shot out of the cannon cuase it has more force and power then the ball dropped...
ok you guys that are questioning eric saying stuff about shooting it stright up are totaly missing the fact that it has to go 4 miles away from the cannon. the only way that they would hit at the same time is if the cannon ball that is dropped and the cannonball shot both start from the same spot and the ball is shot at a degree of 0. then it wouldnt matter how hard you shot the cannon ball both it and the dropped one would hit at the same time.
Well maybe not the only way but definitely the easiest
Rob
All this assumes the ground each hits is level with the ground the cannon is on, and that both hit ground at the same level
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