Question

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

Solution

## You could solve that by squaring the both sides. Instead you could use this method. Explanation: The problem is in 2 parts: The stone falls under gravity to the bottom of the well. The sound travels back to the surface. We use the fact that the distance is common to both. The distance the stone falls is given by: S=(1/2)g(t1)²   -(1) We know that average speed = distance travelled / time taken. We are given the speed of sound so we can say: d=340×(t2)  -(2) We know that: (t1)+(t2)=6.53s We can put (1) equal to (2)⇒ ∴340×(t2)=(1/2)g(t1)²    -(3) t2=(6.53−t1) Substituting this into (3)⇒ 340(6.53−(t1))=(1/2)g(t1)² ∴2220.2−340t1=(1/2)g(t1)² Let g=10 m/s² ∴5(t1)²+340(t1)−2220.2=0 This can be solved using the quadratic formula: X= ( -b±√(b²-4ac) ) ÷ 2a (t1)=(−340±√115600−(4×5×(-2220.2)) ) ÷ 10 = (-340±√160004)/10 = (-340±400.005)/10   Ignoring the -ve root this gives: t1=60/10=6s ∴t2=6.53-6.00=0.53s Substituting this back into (2)⇒ S=340×(t2)=340×0.53=180.2m The answer is 180.2m Hope you have your doubt clearedPhysics

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