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)?
Solution
Use s = ½gt2. So fall time for lagging particle is √(600/g) - √(2·10-6g), giving distance 300 + 10-6 - √(12·10-4). So the distance apart is 20√3 mm = 34.6 mm.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03