Question

Drops of water fall from the roof of a building 18m
high at regular intervals of time. When the first drop
reaches the ground, at the same instant fourth drop
begins to fall. What are the distances of the second
and third drops from the roof?
(1) 6 m and 2 m (2) 6 m and 3 m
(3) 8 m and 2 m (4) 4 m and 2 m​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Height of the building (S) = 18 meters.
• Initial velocity of every drop (u) = 0 m/s.
• g = 10 m/s².
• Let the first drop fall at time "t".

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Applying Second kinematic equation,

$\large{\boxed{\tt S = ut + \dfrac{1}{2}at^2}}$

Substituting the values,

$\large{\tt 18 = 0 + \dfrac{1}{2} \times 10 \times (t)^2}$

Now,

$\large{\tt 18 = \dfrac{1}{\cancel{2}} \times \cancel{10} \times t^2}$

$\large{\tt 18 = 5t^2}$

$\large{\tt t^2 = \dfrac{18}{5}}$

$\large{\tt t = \sqrt{ \dfrac{18}{5}}}$

As we know,

• √18 = 3√2

Now,

$\large{\boxed{\tt t = 3 \sqrt{ \dfrac{2}{5}}}}$

As question said,

The fourth drop leaves when first drop reaches the ground,

Therefore, at a time only 3 drops are in air.

Applying algebra,

$\large{\tt 3 drop = 3 \sqrt{ \dfrac{2}{5}} \: Seconds}$

Therefore,

$\large{\tt 1 drop = 3 \sqrt{ \dfrac{2}{5}} \times \dfrac{1}{3}}$

$\large{\tt 1 drop = \cancel{3} \sqrt{ \dfrac{2}{5}} \times \dfrac{1}{\cancel{3}}}$

It comes as,

$\large{\boxed{\tt 1 drop = \sqrt{ \dfrac{2}{5}} \: Seconds}}$

Note:-

The above got result is equal intervals of time,

After which every drop falls.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

For second drop

$\large{\tt S_1 = \dfrac{gt^2}{2}}$

[This is the simplified form of second kinematic equation ; as u = 0]

Substituting the values,

$\large{\tt S_1 = \dfrac{10}{2} \times \left[2 \times \sqrt{ \dfrac{2}{5}}\right]^2}$

Here we have taken time

as

${\tt t = 2 \times \sqrt{ \dfrac{2}{5}}}$

Because the drop has fallen for two equal intervals of time.

Simplifying,

$\large{\tt S_1 = 5 \times 4 \times \dfrac{2}{5}}$

$\large{\tt S_1 = \cancel{5} \times 4 \times \dfrac{2}{\cancel5}}$

It comes as,

$\large{\tt S_1 = 4 \times 2}$

$\huge{\boxed{\boxed{\tt S_1 = 8 \: meters}}}$

So, the distance of second drop is 8 meters.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

For Third drop

$\large{\tt S_2 = \dfrac{gt^2}{2}}$

[This is the simplified form of second kinematic equation ; as u = 0]

Substituting the values,

$\large{S_2 = \dfrac{10}{2} \times \left[ \sqrt{ \dfrac{2}{5}}\right]^2}$

Here we have taken time

as

${\tt t = \times \sqrt{ \dfrac{2}{5}}}$

Because the drop has fallen for one equal intervals of time.

Simplifying,

$\large{\tt S_2 = 5 \times \dfrac{2}{5}}$

$\large{\tt S_2 = \cancel{5} \times \dfrac{2}{\cancel5}}$

It comes as,

$\large{\tt S_2 = 1 \times 2}$

$\huge{\boxed{\boxed{\tt S_2 = 2 \: meters}}}$

So, the distance of Third drop is 2 meters.

Therefore,

Option - (3)[8 meters and 2 meters]is correct.

$\rule{300}{1.5}$

#refer the attachment for figure.

Answer 2

Answer:

he mate here is your answer

Explanation:

I solve this question using galileo's law.