Question

A stone is dropped from the top of a tower 500m height into a pond of a water at the base of the tower.whennis the slash heard at top​

Answer 1

Answer:

Hey Buddy here's ur answer

11.47 s

Answer 2

$\bold{\huge{\underline{\fbox{\mathfrak{\color{olive}{ello}}}}}}$

$\red{\bold{\underline{\underline{Given\:Data:-}}}}$

$Initial velocity ( <strong>u</strong><strong> </strong>) = 0 m/s$

$Height ( <strong>h</strong><strong> </strong>) = 500 m$

$gravity ( <strong>g</strong><strong> </strong>) = 10 m/s^2$

$Final velocity ( <strong>v</strong> ) = 340 m/s ( speed \:of \:sound \:in \:air)$

$\pink{\bold{\underline{\underline{To\:Find:-}}}}$

Time taken to hear the sound (T) = ?

$\green{\bold{\underline{\underline{Solution:-}}}}$

⟹let the time taken to reach at the base be (t)

⟹and the time taken to reach at the tower from base be (t')

By using 2nd equation of motion

$⟹s = ut + \frac{1}{2} a {t}^{2} \\ \\ ⟹500 = 0 + \frac{1}{2} (10) {(t)}^{2} \\ \\ ⟹500 = 5 {(t)}^{2} \\ \\ ⟹ \frac{500}{5} = {t }^{2} \\ \\⟹ 100 = {t}^{2} \\ \\ ⟹ \sqrt{100} = t \\ \\ ⟹10sec = t$

Now,

$speed \: = \frac{distance}{time}$

$⟹340 = \frac{500}{t'} \\ \\ ⟹340 \: (t') = 500 \\ \\⟹ t' \: = \: \frac{500}{340} \\ \\ ⟹t' = 1.5sec$

Then,

⟹$T = t' + t$

⟹$T = 1.5 sec + 10 sec$

⟹$T = 11.5 sec$

$\bold{\huge{\underline{\fbox{\mathfrak{\color{teal}{Time = 11.5 sec}}}}}}$

$\huge\underline\mathfrak\orange{Thanks}$