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Small steel balls fall from rest through the opening at A at the steady rate of two per second. Find the vertical separation h of two consecutive balls when the lower one has dropped 3 meters. Neglect air resistance.

Respuesta :

Answer:

2.61 m separation

Explanation:

x = xo + vo t + 1/2 at^2      x o   and vo = 0   so this becomes

x  = 1/2 at^2

Find t    when   x = 3

3 = 1/2 (9.81)(t^2)

t = .782 sec

then the SECOND ball is at

x = 1/2 a t^2           but t is reduce by 1/2 second drop rate

x = 1/2 (9.81)(.782 - .5)^2 = .39  m

3 - .39  m = 2.61 m