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.

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

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