Answer to Question #227442 in Mechanical Engineering for yaya

Question

Water drips from a tap at uniform rate onto the floor which is 1.5m below. The first drop strikes the floor at the instant the fourth drop is on the verge of falling from the tap. What would be the location of the different drops when the first drop hits the floor? D1 is on the floor and D4 is at the tap

Answer 1

Given that the first drop hits the floor falling from 1.5

$s= ut+\frac{1}{2}at^2\\ 1.5= 0+\frac{1}{2}*9.8*t^2\\ t= 0.55328s$

Let drops from the top fall after t₁

$t= 3t_1\\ t_1= 0.18442s$

The location of 4th floor = 0 m

The location of 3th floor

$s_3= ut_1+\frac{1}{2}at_1^2\\ s_3= 0+\frac{1}{2}*9.8*0.18442^2\\ s_3=0.16665 m$

The location of 2nd floor

$s_2= ut_1+\frac{1}{2}at_1^2\\ s_2= 0+\frac{1}{2}*9.8*(2*0.18442)^2\\ s_2=0.66661m$

Location of 1st drop is 1.5 m

So, D1 is on the floor and D4 is at the tap.