Use s = ½gt 2 . So fall time for lagging particle is √(600/g) - √(2·10 -6 g), giving distance 300 + 10 -6 - √(12·10 -4 ). So the distance apart is 20√3 mm = 34.6 mm. 7th Eötvös 1900 © John Scholes jscholes@kalva.demon.co.uk 29 Oct 2003 Last corrected/updated 29 Oct 03
7th Eötvös 1900 Problem 3 Two particles fall from rest 300 m under the influence of gravity alone. One particle leaves when the other has already fallen 1 μm. How far apart are they when the first particle reaches the end point (to the nearest 100 μm)?