throwing a ball...

They say the same when you shoot a rifle horinzontal. the bullet will drop the same as if you dropped the bullit from the same height.

OK. Re your rifle bullet fired vs. a dropped bullet at exactlt the same instant. Will they both land at the same time? To take this to the extreme, suppose we have a 16-inch naval rifle which fires a 1 ton shell 18 miles and it impacts the level ground severel seconds later. At the same time that the fired shell leaves the barrel of the gun, an identical 1 ton shell is dropped from the same height as the end of the cannon. Will they land at the same instant (assuming that the curvature of the earth is countered by a rising landscape such that the "ground" is at the same height for 18 miles. Isn't this a vector problem, where one shell has only a verticle vector component, while the other shell has a very significant horizontal vector component? I don't think it would take a second for the "dropped" shell to hit the ground, but the fired shell certainly won't impact in one or even two, three, four,,,, seconds.

This is invariably set as a 1st year dynamics problem. If you're not up with it, I'm not the guy to explain (at this time of night - maybe tomorrow), but be assured, (neglecting air resistance etc. etc.) - they will land at the same time.

As you correctly stated, this is a vector problem - the solution of which will lead you to the conclusion stated above.

[I've just had a look around on the web and I can't believe how hard it is to find out about this elementary stuff. The fired bullet or shell obeys the laws of ballistics - look that up & you'll find a load of stuff about rifle bullet impacts. It follows a parabolic flight path - look that up and you get stuff about "Zero-G". Maybe someone needs to put out an easily findable & understandable article.]

Sorry if I'm being patronising,

John.

[Edit: You have to remember that the naval gun must fire the shell horizontally.]

Sums:

Gun is say, 25m above ground/sea level. Time for shell to drop vertically is found from:

s = u*t + ½*f*t²

where s = distance (metres), u = initial velocity (ms^-1), t = time (s), f = acceleration (ms^-2).

so 25 = 0*t + ½*9.8*t² (9.8 ms^-2 is acceleration due to gravity).

from this, t² = 25/4.9, t ≈ √5.1 ≈ 2.23 seconds.

This is the time of flight. 18 miles ≈ 30,000 m, so using the same equation, this time for the horizontal motion of the fired shell,

30,000 = u*t ( no horizontal acceleration )

u ≈ 30,000/2.23 ≈ 13,000 ms^-1 - that's about 29000 mph, which is the average velocity the shell would have to have.

[Note: this is neglecting air resistance, curvature of the earth etc.etc. This velocity is in fact greater than the escape velocity from Earth, so if you re-introduce curvature (still ignoring air resistance!), the shell will never be seen again. Good gun for shooting down satellites!

Apologies to raj for the formulae.]

_{Edit: I marked this off-topic, but somehow (going from preview & back to edit etc.) the OT flag dissappeared. Sorry, all.}

It would seem that way, but taking all induced lift out of the equation from the volicity of the shell.

Gravity would have the same effect on both shells. As far as the 16 in'er shooting eighteen miles, that gun is not level.

I would have to say that the naval gun is say 30-some odd feet above the seas. I would say both would spash at the same time. (if shot level).

I should have read farther JohnDG says it better.

If the horizontally thrown ball goes REALLY far, it will hit after the vertically dropped ball. The curvature of the earth will result in a longer vertical component of travel. However, you would need to shoot the ball several miles, or measure the time difference in femtoseconds, in order to see any measurable differential.

Does that really work? It occurs to me that vertical refers to the distance directed towards the centre of the earth, ie the radial distance, - in which case, it'll be the same no matter how far it goes (I think - it's getting late!).

Ok, just reconsidered. If it was thrown at (and maintained) orbital velocity ... 'nuff said.

That would make the thrown ball fall faster but hit the ground at the same time if the point of reference was tangential.

You stated in your question that the masses are the same, the answers given would be the same even if the masses where different. Imagine a large lift one person steps into the lift and rides it down, the other runs into the lift, and the lift is large enough for him to continue running, then no matter how fast he runs he get to the ground at the same time. Hope this none formula description helps.

Regards JD.

The more speed the horizontally thrown ball has the later it will touch the ground compared with the vertically dropped ball.
The dropped ball has only a linear vertical component, the other has an additional horizontal which may be much bigger than the other one, depending on the horizontal speed Vo.

No formula wanted? Try with your children, they will have fun !

Regards Uwe

Several people have said that the thrown ball will land later because of the curvature of the earth. In fact this effect depends on which direction you throw the ball: imagine you are at the equator and you fire a bullet at 1000 miles per hour west, then, this is the object which is falling directly towards the centre of the earth whereas the object dropped is the one with a tangential component of velocity.

Of course; now I'm ignoring interaction with the air, which would have a much greater and probably unpredictable effect (especially at this sort of speed).

Air resistance would generate a force proportional to the square of the velocity and in a direction opposite the velocity (ignoring funny effects such as vortex shedding). If you call the vertical component of velocity vy and the velocity of the thrown ball vr, then the upward drag force on the dropped ball will be proportional to vy*vy. The drag force on the thrown ball will be (vr*vr) and its upward component will be (vr*vr)*(vy/vr) or vr*vy. Now, since vr has to be greater than vy, with air resistance the upward drag force on the thrown ball is greater and it hits the ground later.

Unless he throws a sinker where the rotation of the ball is such that it creates a downward force.

The balls will only land at the same time if they also have the same surface area because of air resistance. Your question only stated that they have the same mass and did not state to ignore air resistance, therefore we cannot assume any time that either ball will land.

To illustrate this point, take a sheet of paper and drop it. You'll notice it glides slowly down. Now, crumple that sheet of paper into a ball and drop it from the same height, and you'll see it travels much faster, although its mass hasn't changed.

Of course, if one wants to be picky, your question doesn't even say if the balls we're dropped/thrown from the top of the same building...

If you really want to blow your physics teachers mind (judging by the question and your spelling skills I'm guessing you are a high school student), tell him that there is a non-zero probabillity that both balls are already on the ground at any given time, even before they are released.

<...The balls will only land at the same time if they also have the same surface area because of air resistance....>

Er, what about the length of the path that each ball travels? If air resistance is to be factored into the thinking, then the ball that travels the longer path will have been slowed for longer by drag forces, particularly as it has a non-zero speed at the moment of departure from the top of the building, unlike the ball that is dropped. Intuition suggests therefore that the ball that is dropped vertically will touch the ground first, though there won't be much in it (unless one 'bends it like Beckham', that is).

So there. And not an equation in sight....

We engineering types can easily go off on tangents, and naturally want to qualify statements with various disclaimers.

So, just to be sure you've not been led astray: Both balls hit the ground at the same time. The question (or demonstration) is used to show that the components of motion can be treated independently -- in other words, the fact that the thrown ball is moving horizontally has no effect on its acceleration earthward.

any aeronautical engineer may argue with that

shape does play a roll (if he was throwing something other than a ball). Oops tangent…. Well at lease spin on the ball would effect even if it was applied to both balls.

Or even a knuckle ball, Bob Uecker "voice of the Milwaukee Brewers", once said the best way to catch a knuckle ball is wait till it stops rolling.

Sometimes I like to be difficult, especially on slow uneventful days.

any aeronautical engineer may argue with that

Not any aero engineer... only one who wants to mess with the head of this poor guy who is evidently struggling with both math and the most basic high school physics. There are perhaps 100 reasons for the two balls hitting the ground at slightly different times. But we all know that this guy might soon be reading further in a physics text in which they use this as a central concept in understanding vector addition, projectile motion, etc.

You're all wrong. There is no gravity. The Earth sucks! ;)

The one dropped vertically.

4.3.08

Good crisp reply.We aint not living in vacuum. the flight path of the hori ball will be longer than the verti ball. The "throwing force", if equally delivered in the verti ball, will as it is speed descent.