A stone is dropped into a well 44.1m deep.after how much time the sound will be heard if the velocity of sound is 330m/s
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From second equation of motion
s= ut +1/2gt^2
Given s=44.1m u=0m
v=300m/s
44.1 = 1/2 ×10t^2
8.82 =t^2
About 2.9
s= ut +1/2gt^2
Given s=44.1m u=0m
v=300m/s
44.1 = 1/2 ×10t^2
8.82 =t^2
About 2.9
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