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Mechanical Properties Of Fluids

Question
CBSEENPH11018754

A stone dropped from top of the tower 300 m high splashes into water of pond near the base of tower. When is the splash heard at the top? Velocity of sound in air = 340m/s, g = 9.8 m/s2.

Solution
Let t be the time taken by stone to reach the surface of water in pond.
When the stone strikes the surface of water, splash is produced.
Let, t2 be the time taken by the splash to reach top of tower.
The total time after which the splash is heard after dropping the stone is, 
                       t = t1 + t
The stone is dropped from height 300m.
 The time taken by the stone to reach the surface of water can be calculated by using the thrid equation of motion,  
                       straight h equals ut subscript 1 plus 1 half at subscript 1 squared 
                 space space space 300 equals 1 half cross times 9.8 cross times straight t subscript 1 superscript 2
                       straight t subscript 1 equals square root of fraction numerator 2 cross times 300 over denominator 9.8 end fraction end root space equals space 7.825 straight s 
The time taken by splash to reach the top of tower is,
                        space space space space straight t subscript 2 equals fraction numerator Distance space travelled space by space sound over denominator Velocity space of space sound end fraction 
     equals 300 over 340 equals 0.882 straight s  
Therefore, total time taken to hear the splash at the top is, 
                          
Total time straight t equals straight t subscript 1 plus straight t subscript 2 equals 7.825 plus 0.882 equals 8.707 straight s