# Settle an argument please....

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Hi all, first post - so please be gentle.

I am, by far, no expert on all things scientific - so I'd like some help on this query.

If, say, you dropped a golf ball and a ten pin bowling ball from the Empire State Building simultaneously, on a calm day, would both hit the ground at the same time?

If someone could advise, or post a URL to a definitive answer it would be great!

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Basically yes. gravity applies an acceleration to objects and the acceleration is the same for everything. You would be surprised how many people think gravity applies a speed.

Now if you bring in air resistance then the bowling ball would hit first since the golf ball would experience (relatively) a greater drag force.

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The deceleration from drag on a spherical body (a ball) goes like 1/r2 or 1/r, depending on the nature of air-flow past the ball. In either case, the smaller ball experiences a greater deceleration due to drag. So, the golf ball slows down more and hence, lands second.

Note : insane_alien, the drag force is greater on the bigger ball, but the deceleration is greater on the smaller ball.

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How much of a difference, in terms of time, are we talking here. Is it marginal?

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The deceleration from drag on a spherical body (a ball) goes like 1/r2 or 1/r' date=' depending on the nature of air-flow past the ball. In either case, the smaller ball experiences a greater deceleration due to drag. So, the golf ball [b']slows down more[/b] and hence, lands second.

Note : insane_alien, the drag force is greater on the bigger ball, but the deceleration is greater on the smaller ball.

They both accelerate until terminal velocity is reached then stay at that speed.

The terminal velocity of the golf ball is probably less, due to the "relatively" greater drag (greater for it's weight) as Insane Alien pointed out.

1/r sounds right for deceleration as long as long as the nature of the flow does not change and the density of the sphere is constant.

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Hmmm.. DQW, I thought the Drag coefficient for a sphere was a constant? or are we factoring in those little dimple thingies on the golf ball?

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Hmmm.. DQW, I thought the Drag coefficient for a sphere was a constant? or are we factoring in those little dimple thingies on the golf ball?

IIRC It is around 0.2 for a laminar boundary layer but changes fairly abruptly for a turbulent boundary layer to 0.1. In fact a rougher or dimpled surface, which increases the skin friction, can reduce the total drag considerably at some reynolds numbers.

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How much does the golf ball weigh?

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They both accelerate until terminal velocity is reached then stay at that speed.
Yes, that's correct. I didn't mean to imply that the only force was a decelerating force. Naturally, the accelerating force (mg downwards) was omitted because the acceleration due it it was equal for both balls. So, I was thinking entirely in the downward accelerating frame.

For a sphere (in laminar flow), the terminal velocity goes like [imath]\sqrt{r} [/imath]. But for balls thrown off buildings and such, they will not likely reach terminal velocity.

YT : I didn't factor in the effect of the dimples (which is important); the drag force on a sphere depends on the radius (and density) of the sphere.

The time difference between the landings will depend on the height from which the balls are thrown. For a height of about 5m (or about 16 ft, roughly a second floor window), my rough estimate says that the time difference (on a windless day) will not be more than 40 milliseconds, if the golf ball were perfectly spherical. I can't say anything about the magnitude of the effect of the dimples, though I'm sure this is well documented and must be findable.

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For a sphere (in laminar flow), the terminal velocity goes like [imath]\sqrt{r} [/imath']. .

Sounds right, and also with the square root of the density and inverse square root of the drag coefficient.

Without dimples/roughness on either ball the boundary layer of the bowling ball would become turbulent first, resulting in a drop in the drag corfficient.

How high is the Empire state building? The higher it is the more the bowling ball wins by, I'm sure it is at least as dense as the golf ball and significantly bigger radius.

I think a smooth golf ball would have reached/approached it's terminal velocity but I'm not sure about the bowling ball or a dimpled golf ball.

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To make things harder, I don't think so !

A water balloon, and a plain air balloon, both identical and inflated to the exact same size, will be affected by the same air drag resistance, and will have different weights.

For very sure, launch both from your Empire State building, and you will find that a long time after that pedestrian dries out, the air inflated one is still flying. Even if you do it in your bedroom with zero crosswind, they will not fall at the same speed.

It's the law of selective gravity.

Miguel

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Externet: that is because the drag force on the air filled ballon cancels out the weight(terminal velocity) very quickly so it will only fall at a slow speed. if it was a vacuum and the balloons could hold in the pressure then they would hitthe ground atthe same time.

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To make things harder' date=' I don't think so !

A water balloon, and a plain air balloon, both identical and inflated to the exact same size, will be affected by the same air drag resistance, and will have different weights.

For very sure, launch both from your Empire State building, and you will find that a long time after that pedestrian dries out, the air inflated one is still flying. Even if you do it in your bedroom with zero crosswind, they will not fall at the same speed.

It's the law of selective gravity.

Miguel[/quote']Also, the two balloons have extremely different densities, and hence, the buoyant force on one of them (as a fraction of its weight) is several times the other's. With the golf and bowling balls, the density of both are pretty high and in the same ballpark (no pun intended). The buoyant forces on them are negligible.

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