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A stone is dropped from the top of a tower `500 m` high into a pond of water at the base of the tower. When is the splash heard at the top ? Given, `g = 10 m//s^2` and speed of sound `= 340 m//s`. |
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Answer» Height of the tower, s = 500 m Velocity of sound, `v = 340 m s^(-1)` Acceleration due to gravity,` g = 10 m s^(-2)` Initial velocity of the stone, `u = 0` (since the stone is initially at rest) Time taken by the stone to fall to the base of the tower, `t_(1)` According to the second equation of motion: `s=ut_(1)+(1)/(2)g t_(1)^(2)` `500=0xxt_(1)+(1)/(2)xx10xxt_(1)^(2)` ` t_(1)^(2)=100` `t_(1)=10 s` Now, time taken by the sound to reach the top from the base of the tower, `t_(2)= 500 //340 = 1.47 s` Therefore, the splash is heard at the top after time, t Where,` t= t_(1) +t_(2) = 10 + 1.47 = 11.47 s`. |
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