Question

Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. Find the locations (above the floor in cm) of the second and third drops when the first strikes the floor. Second drop? Third drop?

Answer 1

Using kinematic equations:

Given that drops falls from a height of 200 cm & when 1st drop strikes the floor, then 4th one begins to fall

time taken by first drop to reach the ground will be:

Using 2nd kinematic equation: (Assuming downward is positive)

h = U*t + (1/2)*g*t^2

U = initial speed of drop = 0 m/sec

h = height from which drop falls = 200 cm = 2.00 m

g = 9.81 m/sec^2

So,

2.00 = 0*t + (1/2)*9.81*t^2

t = sqrt (2*2.00/9.81) = 0.63855 sec

Now given that drops are falling at a regular intervals, (time interval between 1st and 4th drop will be 3)

So time between each drop = 0.63855/3 = 0.21285 sec

Now distance travelled by 2nd drop, when first one strikes the ground will be:

h2 = U*t2 + (1/2)*g*t2^2

t2 = 0.63855 - 0.21285 = 0.4257 sec (time for which 2nd drop has travelled)

h2 = 0*0.4257 + (1/2)*9.81*0.4257^2

h2 = 0.889 m = 88.9 cm

location of 2nd drop above the floor will be = 200 - 88.9 = 111.1 cm

Now distance travelled by 3rd drop, when first one strikes the ground will be:

h3 = U*t3 + (1/2)*g*t3^2

t3 = 0.63855 - 2*0.21285 = 0.21285 sec (time for which 3rd drop has travelled)

h3 = 0*0.21285 + (1/2)*9.81*0.21285^2

h3 = 0.222 m = 22.2 cm

location of 3rd drop above the floor will be = 200 - 22.2 = 177.8 cm

Let me know if you've any query.