Physics / Math question
 

At time = 0 a stone is dropped from the top of a cliff above a lake.

Another stone is thrown downward 1.6 seconds later from the same point with an inital speed of 32 m/s.

Both stones hit the water at the same instant. Find the height of the cliff.

E=MC2

:helpme

Remember that the ONLY difference between the two rocks is the initial velocity. one has VI'=0 and other is VI''=32

Accelleration is the same, mass is the same (I assume), distance is the same.

51.2m

(edit: first guess, but wrong)

Google it. Google's supposedly got some kickass calculation software hooked into their engine.

damn i don't remember the old formulas for that, it should be easy if you know them

first do distance = distance
left side of equation is 1/2 g (t-1.6)^2 + 32(t-1.6)
right side is 1/2 g t^2

solve for t

then plug in that value of t to either side to find the value of height


Thats pretty close... not bad for in my head anyway.

into a lake? No way to know... since a lake could be at any level above or below sea level, thus effecting the gravitational pull... altering the speed of the falling rock.

Now... had you said ocean....

not enough info

Is it on earth?

the answer is 27.6 meteres.... I just dont know the equations to use :P

acceleration rate of gravity is 32 fps square over an ocean, lake, or dry land (not taking wind resistance into account)

yeah, you got it, thanks man! :thumbsup

Do they actually teach physics in the USA in imperial units?

That seems very very backward.

84 metres?

WOOO HOOOO.

Cool... Good to know that shit is still in my brain somewhere even though I never use it anymore.

No its always metric from grade school through grad school... at least it was for me.

After careful consideration of all the data, I will say 3.

I don't know about now but back in 1983 they did. :)

damn, 3 seconds late. i just forgot the formula.

Touche.

:thumbsup