A rock is dropped down a dark well and a splash is heard 3 seconds later. Taking into account the time required for sound to travel up the well, calculate the distance to the water in the well. The speed of sound is 340 m/sec.
Hey how do i solve this??? It's driving me nuts!!! Thank u in advance!! asap!
Look familiar at all? What if we substitute Trock with just plain t:
4.905t^2 + 340t - 1020 = 0
It's a quadratic equation! Now you can plug it into the quadratic formula:
t = (-b+/- sqr(b^2-4ac))/ 2a
t = (-340+/- sqr(340^2-4(4.905)(-1020)))/ 2(9.81)
The two answers I got were about -72 and 2.88. Obviously, the stone can't travel back in time, so it must have fallen for 2.88 seconds.
Now we can substitute that answer back into the distance equation for the rock:
Drock = 0.5aTrock^2
= 0.5*9.81*2.88^2
= 40.69
So, your well is 40.69 m deep.
Always check your answers to make sure they make sense. If the well was 1020 metres deep, that's a kilometre deep! Doesn't make much sense, does it? [ NinjaNeer's advice column | Ask NinjaNeer A Question ]
ThirdQED answered Friday August 7 2009, 1:26 pm: We are given that the speed of sound is 340 m/sec.
And the splash is heard 3 seconds later.
Thus:
340 x 3 = 1020 meters
The answer: The distance to the water in the well is 1020 meters (or about 3346.5 feet).
By the way, you've come to the wrong place. This is not a homework-help site, lol. You might want to ask these questions on a different site =) [ ThirdQED's advice column | Ask ThirdQED A Question ]
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