Suppose someone dropped a bowling ball off the Empire State Building, timing it just right so as not to hit anyone below. Would the bowling ball break the concrete, or would the concrete shatter the bowling ball?
Please don’t try this, BTW!!!
I hate to break the illusion, but the Empire State Building is tiered. If you throw things off they land on the 92th floor balcony.
In that case, World Trade Center, same question. :rolleyes:
Why would this necessarily have to be a one-or-the-other case? Chances are most likely that both the bowling ball and the concrete would be heavily damaged.
I’m not certain, but i believe bowling balls are denser than your basic concrete. As a result, the concrete would sustain more damage than the volume of the bowling ball (does this make sense?). However, there’s a lot more concrete out there than there is bowling ball, so the ball would likely disintegrate before too long.
Fire a cannonball at a castle wall and the results will be roughtly the same: cannonballs are a good deal denser than bowling balls, but they still disintegrate. Not before they get to damage the wall though.
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- When I was in eight grade I and some friends had the chance to -uh, gain surrepititious entry to a four-story building under construction. One of the things we did was drop a bowling ball off the fourth-floor balcony onto the pavement of an enclosed patio area. ~ The pavement didn’t break, the bowling ball did. Which isn’t to say that no sidewalk would break, but this one didn’t. The bowling ball cracked away in layers, as if they had made it by making a little marble-sized ball, and then had dipped it into liquid plastic over and over again, adding a bit more size each time. It did not break in half: most of it shattered away into small thin curved pieces, but there was still a softball-sized center left intact but with a fairly uneven surface. You could see the bottoms of the finger holes, but you couldn’t pick it up by them anymore. Not that you had to. - MC
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That’s what we need around here, more epirical data. :D:D
From physics classes, we know that 1/2 * mass * (square of speed) = mgh (kinetic energy equal to height energy). Thus we get speed = square root of ( 2gh), and with the Emp. State Buliding having a height h of -damned I read it a thousand times but I have to look it up- 381 metres, you get a falling speed of about 86 metres per second or 195 miles per hour. I don’t know how the concrete likes that, but I suppose if you shoot your bowling ball with 195 mph onto concrete, both of them will not look too good afterwards.
Why do you come to the conclusion that the density is the deciding factor? Density is a measure of mass per volume; it is not a measure of tensile or compressive strength, nor of toughness. By saying this, you are saying the strength of the concrete and the ball do not really matter.
Here’s how the process will work:
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The bowling ball will accelerate towards the concrete below until it reaches terminal velocity. If I had my fluids book here I could calculate the terminal velocity. I tried many, many web searches, and was amazed that nowhere can I find the terminal velocity of a bowling ball.
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The bowling ball will acquire a certain kinetic energy. Here is where the density sort of comes into play, as it determines for any volume of bowling ball the mass of the ball. And kinetic energy is 0.5massvelocity[sup]2[/sup]. This is energy that must be dissipated if the ball is to be brought to rest.
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The ball impacts the concrete with a force. The concrete will do several things at this point. It will deform under the stress of having to decelerate the ball. If the strength of the concrete and it’s reinforcing steel is not great enough, it will crack, break, or shatter. Meanwhile, the ball is doing the same thing. It will deform at first, then depending on it’s strength it will crack, break, or shatter explosively.
So the density influences the kinetic energy that must be removed, but it has nothing directly with the strength of the concrete or the bowling ball.
Right… sounds like you know a lot more about this than i do. Sorry, i should have mentioned my post was a WAG.
Hey, don’t be sorry. 
I really just wondered if you were simply trying to explain the problem in a different way, or something. I certainly don’t have all the answers about engineering or physics - unless it’s about coal.