Question

A tap drops water at one second interval .find the distance between 3rd and 5th drop when 6 th drop is just coming out of tap ?​

Answer 1

Answer:

Take the case of Ist drop.

Let the time it takes to touch the ground, i.e. to move a distance of 5 m is t.

Then, s=1/2gt^2

5=1/2×10×t^2

t=1 sec.

Therefore, time interval is 1/2 sec.

That is,after every 1/2 sec.,a drop is released.

By the time 3rd drop is released, 2nd drop utilises 1/2 of the second.

S=1/2×g×t^2=1/2×10×1/4

=5/4=1.25 m

Answer 2

AnSwer :-

Given :-

⟶ Initial velocity, u = 0m/s

⟶ Accerlation due to gravity = -10m/s²

⟶ Time interval between 2drops = 1sec

To find :-

The distance between 3rd and 5th drop

Solution :-

The distance between 3rd and 5th drop = distance between 6th and 3rd drop - distance between 4th and 5th drop

so,

let us find distance between 6th and 3rd

let the distance be 's'

using equation of motion

.i.e.,

$\longrightarrow \sf s = ut + \dfrac{1}{2} a {t}^{2}$

$\longrightarrow \sf s = 0(3) + \dfrac{1}{2} ( - 10)(3) {}^{2}$

$\longrightarrow \sf s = \dfrac{1}{2} ( - 90)$

$\longrightarrow \boxed{ \sf s = - 45m}$

thus distance between 6th and 3rd drop is -45m

Now,

the distance between 6th and 5th drop

let the distance be 's$_{1}$'

using the equation of motion

.i.e.,

$\longrightarrow \sf s_{1} = ut + \dfrac{1}{2} {at}^{2}$

$\longrightarrow \sf s_{1} = 0(1) + \dfrac{1}{2} ( - 10)(1) ^{2}$

$\longrightarrow \sf s_{1} = 0 + \dfrac{1}{2} ( - 10)$

$\longrightarrow \boxed{ \sf s_{1} = - 5m}$

the distance between 6th and 5th drop is -5m

The distance between 3rd and 5th drop

= s - s$_{1}$

= 45m - 5m

= 40m.

Thus, the distance between 3rd and 5th drop is 40m.