A rock is dropped into an abandoned mine, and a splash is heard 4 seconds later. Assuming that sound takes a negligible time to travel up the mineshaft, what is the depth of the shaft and how fast is the rock falling when it hit the water?

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Answer 1 · 2018-06-06T21:34:21

Using some Newtonian Kinematics equations, we find that the distance is 78.4 meters and the rock's velocity was 39.2 m/s

We can use two different equations to find our answers:

#d=V_it+1/2at^2#

#V_f=V_i+at#

The first describes the distance, and the second the velocity. We know that the time #t=4s#, and the acceleration due to gravity is #~=9.8m/s^2#.

Since the rock was dropped and not thrown, we can assume that the initial velocity #V_i=0#.

Let's plug in those values:

#d=(0)(4)+1/2(9.8)(4)^2#

#d=0+4.9xx16#

#color(green)(d=78.4m)#

#V_f=(0)+(9.8)(4)#

#color(green)(V_f=39.2)#

Answer 2 · 2018-06-07T00:04:03

a. #y = 78.4 m#
b. #v = 39.2 m/s#

Ben's answer is fine. this is an slightly different approach.

Calculate the answer to b first.

b. #v = u + g*t#

#v = 0 + 9.8 m/s^cancel(2)*4 cancel(s) = 39.2 m/s#

a. Since the acceleration was uniform (meaning constant), we can use the average velocity approach.

#y = ((u+v)/2)*t#

#y = ((0 + 39.2 m/s)/2)*4 s = 19.6 m/s*4 s = 78.4 m#

I hope this helps,
Steve