Question

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.

Answer 1

2.61 m separation

Step-by-step 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