Asked by anthony
I posted this quesiotn and got some help but still don't know how to do it. could i please get a little more assistance please. I don't know how to solove for the time of falling or sound without the variable height.
eng physics a rock is dropped at sea cliff at the sound of it striking the ocean if head 3.4 seconds later. If the speed of sound is 340m/s how high is the cliff?
3.4 seconds= time for rock to fall+time for sound to come up.
falling: h=1/2 gt^2, or t= sqrt2h/g
sound: h=vsound*t or t=h/speedsound
3.4=sqrt 2h/g + h/speedsound
looks like a quadratic equation to me.
let u^2=h
u^2/speedsound+u sqrt2/g - 3.4=0
solve for u, then go back and solve for h.
eng physics a rock is dropped at sea cliff at the sound of it striking the ocean if head 3.4 seconds later. If the speed of sound is 340m/s how high is the cliff?
3.4 seconds= time for rock to fall+time for sound to come up.
falling: h=1/2 gt^2, or t= sqrt2h/g
sound: h=vsound*t or t=h/speedsound
3.4=sqrt 2h/g + h/speedsound
looks like a quadratic equation to me.
let u^2=h
u^2/speedsound+u sqrt2/g - 3.4=0
solve for u, then go back and solve for h.
Answers
Answered by
bobpursley
multiply by speed of sound
u^2+u*speedsound*sqrt2/g -3.4speedsound=0
This is a quadratic in the form of
au^2+bu+c=0
solve for u.
After you get that, remember h=u^2
u^2+u*speedsound*sqrt2/g -3.4speedsound=0
This is a quadratic in the form of
au^2+bu+c=0
solve for u.
After you get that, remember h=u^2
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