Solving Freefall Problem With Special Attention to Units

In summary, a man standing on a cliff dropped two stones, one after another, and needs to know how long it will take for the distance between the two stones to be 10 meters. Using the formula for distance and the given acceleration due to gravity, it can be solved by setting the distances equal to each other and solving for time. However, when including the given units, the solution does not result in a unit of seconds, indicating a mistake in the calculations.
  • #1
bujorn
3
0

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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  • #2
The t and t-1 have units too: they have units of seconds. Try including those and see what you get.
 
  • #3
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.
 
  • #4
any other ideas?
 

1. How do you solve a freefall problem?

To solve a freefall problem, you need to use the equation h = 1/2gt^2, where h represents the height, g is the acceleration due to gravity (usually 9.8 m/s^2), and t is the time. Plug in the known values and solve for the unknown variable.

2. What units should be used when solving a freefall problem?

When solving a freefall problem, it is important to use consistent units throughout the calculation. Common units of measurement for freefall problems include meters for height, seconds for time, and meters per second squared for acceleration due to gravity.

3. Why is it important to pay special attention to units when solving freefall problems?

Units are an essential part of any scientific calculation. In freefall problems, using the wrong units or mixing units can lead to incorrect solutions. Paying special attention to units ensures that the final answer is accurate and meaningful.

4. Can you solve a freefall problem without using the equation h = 1/2gt^2?

Yes, there are other equations that can be used to solve freefall problems, such as v = gt, where v represents the final velocity and g is the acceleration due to gravity. However, the equation h = 1/2gt^2 is the most commonly used and is often the easiest to work with.

5. Are there any real-world applications of solving freefall problems?

Yes, freefall problems are commonly used in physics and engineering to predict the motion of objects in freefall, such as a skydiver or a falling object. They are also used in designing amusement park rides and studying the motion of celestial bodies in outer space.

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