Question

A stone is dropped from rest the well is 50 m deep how long will it take to reach the bottom

Answer 1

Answer:

${\bold{\therefore Time\:taken=\sqrt{10}\:sec}}$

Explanation:

${\bold{\huge{\underline{SOLUTION-}}}}$

• In the given question information given about a stone dropped from top of the well and height of well is given.

• Initial velocity of stone is 0 and acceleration is in downward direction.

• we have to find the time taken to reach the stone at the bottom.

$\underline \bold{Given : } \\ \implies Onitial \: velocity(u )= 0 \\ \implies Acceleration(a) = g = 10 {m} /s ^{2} \\ \implies Height \: of \: well(s) = 50m \\ \\ \underline \bold {To \: Find : } \\ \implies Time \: taken(t) = ?$

• According to given question :

$\bold{Second \: equation \: of \: motion : } \\ \implies s = ut + \frac{1}{2}a {t}^{2} \\ \implies 50 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ \implies 50 = 5 {t}^{2} \\ \implies {t}^{2} = 10 \\ \bold{\implies t = \sqrt{10} \: sec} \\ \\ \bold{\therefore \: Time \: taken \: to \: reach \: bottom = \sqrt{10} \: sec }$

Answer 2

ANSWER:-

Given:

A stone is dropped from rest the well is 50m deep.

To find:

How long will take to reach the bottom of the well?

Solution:

⚫Acceleration,(a) = 9.8m/s²

⚫Initial velocity, (u)= 0

⚫Distance, (s)= 50m

Therefore,

By using Newton's law of motion;

=) v² - u² = 2as

=) v² - 0² = 2× 9.8× 50

=) v² = 980

=) v= √980

=) v= 31.30

Now,

We got the final velocity v.

So, we can find the time required by Newton's equation of motion;

=) v= u+ at

=) 31.30 = 0 + 9.8× t

=) 31.30= 9.8t

=) t = 31.30/9.8

=) t = 31.30× 100/9.8× 100

=) t = 3130/980

=) t = 3.193 m/sec.

Hence,

3.193 m/s will take to reach the bottom of the well.

Hope it helps ☺️