Solving Freefall Problem With Special Attention to Units

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Homework Statement



A man was standing on a cliff when he dropped a stone. One second later, he dropped another stone. How long before the distance between the two stone is 10 meters? (Show solutions with special attention to deriving the units.) Use g=10 m/s^2.

Homework Equations



Let d_1 = depth of first stone
Let d_2 = depth of second stone
t_1 = time of first stone
t_2 = time of second stone

d_1 - d_2 = 10

d = (1/2)gt^2

t_1 = t
t_2 = t-1

The Attempt at a Solution



I usually solve problems like this but I really have not taken special attention how the units are derived. Since I know that the solving t would result in a unit of seconds, I neglect the units and continue to work on with the problem.

d_1 - d_2 = 10

Substituting the formula for d in d_1 and d_2:

(1/2)10t^2 - (1/2)10(t-1)^2 = 10
5t^2 - 5(t^2-2t+1) = 10
5t^2 - 5t^2 + 10t - 5 = 10
10t - 5 = 10
10t = 10 + 5
10t = 15
t = 15/10 or 1.5 sec

I assumed that the equation is correct, thus t would result in unit of seconds.

However, when I tried to solve the problem including the given units, I ended up like this:

d_1 - d_2 = 10m
(1/2)(10m/s^2)t^2 - (1/2)(10m/s^2)(t-1)^2 = 10m
(5m/s^2)(t^2) - (5m/s^2)(t^2-2t+1) = 10
5mt^2/s^2 - 5mt^2/s^2 + 10mt/s^2 - 5m/s^2 = 10m
10mt/s^2 - 5m/s^2 = 10m
(2t-1)5m/s^2=10m
2t - 1 = (10ms^2)/5m
2t - 1 = 2s^2
2t = 2s^2 + 1
t = (2s^2 + 1)/2

Where did I go wrong? Am I missing something?
 
Last edited:
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ideasrule said:
The t and t-1 have units too: they have units of seconds. Try including those and see what you get.

Thanks for the idea, ideasrule (pardon the pun) :D

Here it goes:

d_1 - d_2 = 10m
(1/2)(10m/s^2)(ts)^2 - (1/2)(10m/s^2)[(ts-1s)]^2 = 10m
(5m/s^2)(t^2 s^2) - (5m/s^2)(t^2- 2t+ 1)(s^2) = 10m
5mt^2 - (5m)(t^2- 2t + 1) = 10m
5mt^2 - 5mt^2 + 10mt - 5m = 10m
10mt - 5m = 10m
10mt = 10m + 5m
10mt = 15m
t = 15m/10m
t = 1.5 (unitless)

Ideasrule, still it did not produce a unit in seconds.