PHYSICS: You drop a stone into a deep well and hear it hit the bottom 3.80 s later.

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PHYSICS: You drop a stone into a deep well and hear it hit the bottom 3.80 s later.

[From: ] [author: ] [Date: 12-09-13] [Hit: ]

5 g td^2= 3.8v- (v x td)Plugging in the values for g and v= 343 ms-1 gives:-4.905 td^2 = (3.8 x 343) - 343td4.905 td^2 =- 343td + 1303.44.......

ts = 3.8 -td

Now, we plug this into (-4-) giving us:-

0.5 g td^2 = v x (3.8-td)
expanding the rhs gives:-

0.5 g td^2 = 3.8v- (v x td)

Plugging in the values for g and v= 343 ms-1 gives:-

4.905 td^2 = (3.8 x 343) - 343td
4.905 td^2 = - 343td + 1303.4

4.905 td^2 +343td - 1303.4=0
Which is a quadratic equation
Where: a = 4.905; b = 343 and c = -1303.4

(-b +/- squar root of [b^2 - 4 x a c])/2a
(-343 +/- sqrt [117649 +25572.7])/9.81
(-343 +/- 378.4)/9.81

td = (-343 + 378.4)/9.81 and (-343 - 378.4)/9.81
td = -35.5/9.81 and -721.4/9.81
td = 3.6 seconds or and -73.5 seconds

td can't be negative, so td = 3.6 seconds

Putting this into (-1a) gives
s = 0.5 x 9.81 x 3.6^2
s =4.905 x 13.03
s = 63.9m
s = 64m

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The 0.5 is just my way of writig 1/2.
The original equation of motion is:

s = ut + 1/2 a t^2 and this can be found in any physics textbook. So, we can also write-
s = ut + 0.5 a t^2

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time to fall >> t = 3.8 - X/343
distance X = 1/2*a* t^2
substitute for t and solve
X=1/2*a*(3.8 - X/343)^2

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