A stone is dropped into a well 19.6m deep and the impact pf sound is heard after 2.056 seconds. Find the velocity of sound in air
t=t1+t2t=t_1+t_2t=t1+t2
h=gt12/2→t1=2h/g=2⋅19.6/9.81≈1.999 (s)h=gt_1^2/2\to t_1=\sqrt{2h/g}=\sqrt{2\cdot 19.6/9.81}\approx1.999\ (s)h=gt12/2→t1=2h/g=2⋅19.6/9.81≈1.999 (s)
t2=h/vt_2=h/vt2=h/v
2.056=1.999+19.6/v→v=19.6/0.057≈344 (m/s)2.056=1.999+19.6/v\to v=19.6/0.057\approx344\ (m/s)2.056=1.999+19.6/v→v=19.6/0.057≈344 (m/s) . Answer
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