Tinku dropped a stone into a very deep well. He hears the sound 6.53 s after he dropped the stone. Find the depth of the well. Take acceleration due to gravity, g =10 m/s² and vsound= 340 m/s.

In the above question how to solve equation √D/5 + D/340 = 6.53

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Solution

√(D/5) + D/340 = 6.53
√(D/5)^2=((6.53)-(D/340))^2
D/5=(42.6409)+(D^2/115600)-(D0.0384)
0.2384D=42.6409+(D^2/115600)

D=180m

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Q. Tinku dropped a stone into a well of depth

180 m

at time

t = 0

. He heard a sound after time

t = x

. What is the value of

x

. Take acceleration due to gravity,

g = 10 m s^{- 2}

and

v_{s o u n d} = 340 m s^{- 1}

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Tinku dropped a stone into a very deep well. He hears the sound 6.53 s after he dropped the stone. Find the depth of the well. Take acceleration due to gravity, g =10 m/s2 and vsound = 340 m/s. In the above question how to solve equation √D/5 + D/340 = 6.53

√(D/5) + D/340 = 6.53 √(D/5)^2=((6.53)-(D/340))^2 D/5=(42.6409)+(D^2/115600)-(D*0.0384) 0.2384*D=42.6409+(D^2/115600) D=180m

Tinku dropped a stone into a very deep well. He hears the sound 6.53 s after he dropped the stone. Find the depth of the well. Take acceleration due to gravity, g =10 m/s² and vsound= 340 m/s.

In the above question how to solve equation √D/5 + D/340 = 6.53

√(D/5) + D/340 = 6.53
√(D/5)^2=((6.53)-(D/340))^2
D/5=(42.6409)+(D^2/115600)-(D0.0384)
0.2384D=42.6409+(D^2/115600)

D=180m