Solving Freefall Problem With Special Attention to Units

  • Thread starter Thread starter bujorn
  • Start date Start date
  • Tags Tags
    Freefall Units
Click For Summary

Homework Help Overview

The problem involves analyzing the motion of two stones dropped from a cliff at different times, specifically focusing on the distance between them after a certain duration. The subject area pertains to kinematics and free fall, with an emphasis on unit consistency in the calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the time at which the distance between the two stones is 10 meters, while also focusing on the proper handling of units in their calculations. Some participants question the treatment of time variables and suggest including units explicitly in the equations.

Discussion Status

Participants are actively exploring different approaches to incorporate units into the problem. There is a recognition of the need to clarify the role of time in the equations, but no consensus has been reached on the correct method or outcome. The discussion remains open with further inquiries for additional ideas.

Contextual Notes

The original poster expresses uncertainty about their calculations and the treatment of units, indicating a potential gap in understanding how to maintain unit consistency throughout the problem-solving process.

bujorn
Messages
3
Reaction score
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:
Physics news on Phys.org
The t and t-1 have units too: they have units of seconds. Try including those and see what you get.
 
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.
 
any other ideas?
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
1K
Replies
3
Views
1K