Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Water drops are falling from a nozzle of a shower onto the floor, from a height of $9.8 \,m$. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

JEE MainJEE Main 2021Motion in a Straight Line

Solution:

image
$ H =\frac{1}{2} gt ^{2} $
$\frac{9.8 \times 2}{9.8}= t ^{2}$
$ t =\sqrt{2}\, sec$
$\Delta t$ : time interval between drops
$h =\frac{1}{2} g (\sqrt{2}-\Delta t )^{2}$
$0=\frac{1}{2} g (\sqrt{2}-2 \Delta t )^{2}$
$\Delta t =\frac{1}{\sqrt{2}}$
$h =\frac{1}{2} g \left(\sqrt{2}-\frac{1}{\sqrt{2}}\right)^{2}$
$=\frac{1}{2} \times 9.8 \times \frac{1}{2}=\frac{9.8}{4}=2.45\, m$
$H - h =9.8-2.45$
$=7.35 \,m$