Question

A stone is released by a miner into a mine 176.4 m deep. Find the time elapsed just before hearing the stone striking the bottom of the mine. (Speed of sound in air=352.8 m/s. g=9.8 m/s?]

Answer 1

✯✯ QUESTION ✯✯

A stone is released from a balloon that is descending at a constant speed of 10 m/s.Neglecting air resistance, after 20 s the speed of the stone is (g = 9.8 m/s2)

(a) 2160 m/s

(b) 1760 m/s

(c) 206 m/s

(d) 196 m/s

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✰✰ ANSWER ✰✰

$\implies\tt{Initial\:Velocity(u)=10m/s}$

$\implies\tt{Acceleration(a)=9.8{m/s}^{2}}$

$\implies\tt{Final\:Velocity(v)=?}$

$\implies\tt{Time\:Taken(t)=20s}$

➥Now ,

➥Using 1st Equation of Motion : -

$\implies\tt{\small{\boxed{\bold{\bold{\green{\sf{v=u+at}}}}}}}</p><p>$

➥Putting Values : -

$\implies\tt{v=10+\dfrac{98}{\cancel{10}}\times{\cancel{20}}}$

⟹v=10+98×2

$\implies\tt{v=10+196}</p><p>$

$\purple\longmapsto\:\large\underline{\boxed{\bf\green{v}\orange{=}\pink{206m/s}}}$

➮So , Final Velocity of the stone is 206m/s.

☛Option (c) is correct....