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.When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?

Answer 1

Answer:

(A) 88.92 cm

(B) 22.22 cm

Explanation:

distance (s) = 200 cm = 0.2 m

initial velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the time (T) it takes for the first drop to strike the floor

from  s = ut + $0.5at^{2}$

         200 = 0 + $0.5 x 9.8 x T^{2}$

         200 = $4.9 x T^{2}$

         200 / 4.9 = $T^{2}$

         T = 6.4

(A) When the first drop strikes the floor, how far below the nozzle is the second drop.

we can find how far the second drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

• s = distance
• u = initial velocity = 0
• t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground ($\frac{2}{3}.T$)
• a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ($0.5 x 9.8 x (\frac{2}{3}T)^{2}$)

s = 0 + ($0.5 x 9.8 x (\frac{2}{3} x 6.4)^{2}$)

s = 88.92 cm

(B) When the first drop strikes the floor, how far below the nozzle is the third drop.

we can find how far the third drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

• s = distance
• u = initial velocity = 0
• t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground ($\frac{2}{3}.T$)
• a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ($0.5 x 9.8 x (\frac{1}{3}T)^{2}$)

s = 0 + ($0.5 x 9.8 x (\frac{1}{3} x 6.4)^{2}$)

s = 22.22 cm