Thread: falling object

A stone ks dropped from a high bridge into the water. A camera takes a photograph every 0.4 seconds showing how far the stone has falle. The equation is given y= 4.9x^2 IF the bridge is 11 meters above the water, how long will it take for the stone to hit the water. I came up with 4.5176 sec. Also after 3 seconds it asks how far has it fallen, I came up with 44.1 meters. The other part asked how far the stone fell in the last hundredth of a second before hittiong the water. I came up with .45 sec. It also asks how fast was the stone traveling before it hit the water? And how fast was the stone traveling when it hit the water. I don't know how to do that part.
Any help would be appreciated!
Thank you very much!!!!!!!
Keith

Originally Posted by keith

A stone ks dropped from a high bridge into the water. A camera takes a photograph every 0.4 seconds showing how far the stone has falle. The equation is given y= 4.9x^2 IF the bridge is 11 meters above the water, how long will it take for the stone to hit the water. I came up with 4.5176 sec. Also after 3 seconds it asks how far has it fallen, I came up with 44.1 meters. The other part asked how far the stone fell in the last hundredth of a second before hittiong the water. I came up with .45 sec. It also asks how fast was the stone traveling before it hit the water? And how fast was the stone traveling when it hit the water. I don't know how to do that part.
Any help would be appreciated!
Thank you very much!!!!!!!
Keith

Ummm... I just LOVE translating Math equations into Physics. NOT!

Just for the sake of argument, I want you to note a few things about the equation y = 4.9x^2.
1) The origin of the coordinate system we are using is at the bridge level.
2) The initial velocity of the object is 0 m/s.
3) The positive direction is downward.
4) x is being used as a time coordinate.

With that all said, let's go for it!

If the bridge is 11 m above the water, we are looking for an x such that y = 11 m (the time the rock hit the water.) So
11 = 4.9x^2

x = sqrt{11/4.9} s = 1.4983 s

After 3 s how far has it fallen? Simply
y = 4.9(3)^2 m = 44.1 m
as you said.

How far did it fall in the last 0.01 s of its fall?
We calculate the position at sqrt{11/4.9} s - 0.01 s = 1.4883 s and the position at sqrt{11/4.9} s.
y(1.4883 s) = 4.9*(1.4883)^2 m = 10.8537 m
y(1.4983 s) = 11 m <-- This one we already knew.

So the stone fell 11 m - 10.8537 m = 0.146343 m
or about 0.146 m in the last 0.01 s.

How fast is the stone travelling just before it hits the water? Velocity is the time derivative of position (displacement) so:
y= 4.9x^2
y' = 9.8x

So at x = 1.4983 s the velocity is
y' = 9.8(1.4983) = 14.6833 m/s

And is your prof trying to play mind games with you? The speed just before it hits the water vs how fast it's going when it hits the water are the same. (Within any physically measurable tolerances.)

-Dan

Thanks for your help! My prof is a little tough, I try hard not to laugh when we get our assignmet each week. I think it usually takes about 30 pages of explainations, a lot of busy work.
Thank you very much for your help!
Keith