Question

Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall.
Find the location of the individual drops when a drop strikes the floor.

Answer 1

Answer:

0.91437 m

0.22859 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

$s=81\ inches=81\times 0.0254=2.0574\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.0574=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.0574\times 2}{9.81}}\\\Rightarrow t=0.64764\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.64764}{3}\\\Rightarrow t'=0.21588\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.21588\\\Rightarrow t''=0.43176\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.43176^2\\\Rightarrow s=0.91437\ m$

Distance from second drop is 0.91437 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.21588^2\\\Rightarrow s=0.22859\ m$

Distance from third drop is 0.22859 m