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A stone is dropped from the top of a tower 500m height into a pond of water at the base of the tower. When is the splash heard at the top ? Given g=10ms^(-2) and speed ofsound =340m//s. |
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Answer» Velocity of sound , `upsilon=340ms^(-1)` Acceleration due to GRAVITY, `g=10m//s^(2)`. Initial velocity of the stone, `u=o` (since the stone is initially at rest) TIME TAKEN by the stone to fall to the baseof the tower, `t_(2)` According to the second EQUATION of motion. `S=ut_(1)+1//2 x 10x t_(1)^(2)` `t_(1)^(2)=100 "" t_(1)=10S` Now, time taken by the sound to reachthe TOP from the base of the tower, `t_(2)=500//340=1.47`s Therefore, the splashis heard at the top after time, t Where, `t=t_(1)+t_(2)=10+1.47=11.47S`. |
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