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twistedps · Nov 7, 2006, 05:53 PM

Hi, I am having a lot of trouble finding out a formula for this problem:
a baseball falling takes 4.0s to go the last 20meters before it smashes into the ground. What is the total height from which the baseball has fallen?

I thought it would be something similar to y = Vot + 1/2 a*t^2
which would be y = (-4.9m^2) * 4.0s but I don't KNOW this isn't correct to get the total height from which it was dropped...
can anyone give me any advice on this?

Capuchin · Nov 8, 2006, 02:22 AM

That's one very slowly falling baseball.. dropping from 0 m/s velocity it should go 78.4 m in 4 s. Is there other information? Are we considering viscous drag? Is this ball falling on Earth? It is being thrown upwards?

you need to work out how fast the ball hits the ground, and from that how far it has fallen under constant acceleration, but with g=9.8, your question doesn't make sense to me.

dmatos · Nov 8, 2006, 05:01 PM

Don't assume that Vo is zero, there's nothing in the question that implies it is. In fact, the question implies that Vo is not zero.

Use your equation y = Vot + 1/2at^2. You have all of the information except for Vo. Then, all you have to do is figure out how far the ball must have fallen under g to attain Vo.

Capuchin · Nov 9, 2006, 03:43 AM

I think my point is that even if you dropped it from 0 m/s, it would travel 78.4 m in 4 s, so it's impossible for an object, under earth's gravity, in a non-viscous situation to fall only 20 m in 4 s. Now if you threw it upwards 10m and it came down 10m, that might take longer, but 1) it'd be about 2 s and 2) I wouldn't call it falling 20 m.

How about if we threw it up to a height of 20 m, then the total upward and downward journey time would approach 4 s, but that isn't falling 20 m, is it? It's freefalling within a 20 m range. In that case the answer would be 0m "a ball falling from 0 m would take 4 s to travel the last 20 m before hitting the ground (as long as it was propelled upwards at 19.6 m/s)"

Please say I've misunderstood something here.

The question confuses me.

You're right that that's how you would get the right answer, but I don't see how the answer would make sense.