Question

A stone is dropped down a deep well from rest. The well is 50 metre deep . How long will it take to reach at the bottom of the well?

Answer 1

At the top of the well u=0 the using s=ut+1/2at^2 s=*0*t +1/2at^2 s=1/2*10*t^2
50=5*t^2 so t=root 10 sec

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

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

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• 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.

• 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.

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

• According to given question :

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