Question

A water droplet is released from a cloud and hit a roof which is inclined 60 degrees to horizontal, after hitting the roof droplet becomes rest. Volume of drop is 0.25 cm3 . Droplet initially drops a distance of 20m under decreasing acceleration and then the rest with uniform velocity, find the height to the cloud if droplet moves 98 seconds under constant velocity

Answer 1

We can divide this motion of the water droplet into 2 parts:

1. Dropping at a distance of 20m under decreasing acceleration.

2.Uniform velocity

Analysis of the first case

We can get the time of the fall.

$S=ut+ \frac{1}{2} g t^2$

Where S is the displacement from the clouds, u is the initial velocity, t is the time.

$20=0 \times t+ \frac{1}{2} \times 9.8 \times t^2$

$t= \sqrt{\frac{20}{0.5 \times9.8}}=2.02 s$

Analysis of the second case

We can find the velocity from the first part, this will be the initial velocity of the second part

$u=st=20 \times2.02=40.41 m/s$

We can now find the height to the roof

$S=u \times t+ \frac{1}{2} \times g \times t^2$

$S=40.41 \times 98+ \frac{1}{2} \times 9.8 \times 98^2=51019.78m$

The height to the cloud , $51019.78 m+20 m=51019.78 m$