How Can You Calculate the Speed of an Object Dropped from a Building?

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SUMMARY

The discussion focuses on calculating the speed of an object dropped from a building, modeled by the equation s(t) = 5t² for 0 ≤ t ≤ 5 seconds. The height of the building is determined to be 125 meters, and the average speed over the entire drop is calculated to be 25 m/s. Participants clarify that the average speed can be computed using the formula v = s/t, despite initial confusion regarding its correctness. The discussion emphasizes the importance of correctly interpreting the distance function and calculating speeds at specific time intervals to approximate instantaneous speed.

PREREQUISITES
  • Understanding of kinematic equations, specifically s(t) = 5t².
  • Knowledge of average speed calculation using v = s/t.
  • Familiarity with graphing functions on a graphics calculator.
  • Basic principles of physics related to free fall and motion.
NEXT STEPS
  • Learn how to derive instantaneous speed from average speed using limits.
  • Explore the concept of acceleration due to gravity and its effects on falling objects.
  • Study the use of graphical calculators for visualizing motion equations.
  • Investigate the differences between average speed and instantaneous speed in physics.
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding motion and speed calculations in real-world scenarios.

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Homework Statement


An item is dropped from a building and its equation is given by s(t)=5t^2
0<(or equal to)t<(or equal to)5.

1.)Draw a graph of this.

2.)Find the height of the building.

The person performing the experiment wishes to calculate the speed at which it hits the ground. He decides he will try Velocity=distance/time.

3.) Calculate the average speed on the item's journey, strictly using v=s/t (even though it is wrong)

He identifies that his formula ism't correct and so attempts to find the speed at each time interval of t=0 and t=1, then t=1 and t=2, and so on up to the between t=4 and t=5

4.) What is the found speed for the intervals.


Homework Equations


I think most of this would be done on Graph mode, on a graphics calculator.


The Attempt at a Solution



I got 125m for the height, 25ms-1 for the overall journey, although i can't remember my answers for the next questions. For the time intervals, i entered Y-cal for the height and entered the distance between each time interval and divided it by the second in between-I'm not sure about this.

Thankyou for your help. This is very easy, but it's hard to do something that leads you through doing the task incorrectly. Basically in these investigations you have to follow the instructions, and in this case you have to follow the student doing the wrong formulas, etc. Although as the paper unfolds the student finds the correct formula.
 
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Well, for 4), he measures the distance covered in each second. Say, then, between the second and first second. The distanced traveled will be s(2) - s(1), and similarly for the other ones.
 
hds29ka said:

Homework Statement


An item is dropped from a building and its equation is given by s(t)=5t^2
0<(or equal to)t<(or equal to)5.

1.)Draw a graph of this.

2.)Find the height of the building.

The person performing the experiment wishes to calculate the speed at which it hits the ground. He decides he will try Velocity=distance/time.
It would be very helpful if you were to write the problem just as it is given. "An item is dropped from a building and its equation is given by s(t)=5t^2 0<(or equal to)t<(or equal to)5." doesn't make sense. "It" refers to the "item" and an "item" doesn't have an equation! Presumably this is supposed to be "the equation of its height is" except that also doesn't make sense- the "dropped item" would be going up! I have to conclude that the given equation is for the distance (In what units? You use meters later- it would have been nice to tell us that initially) below the top of the building after time t (apparently in seconds. Again you did not say that as part of the problem). Finally, there is nothing said about the "0\le t\le5" meaning that the item hit the ground in 5 (seconds).

3.) Calculate the average speed on the item's journey, strictly using v=s/t (even though it is wrong)

He identifies that his formula ism't correct and so attempts to find the speed at each time interval of t=0 and t=1, then t=1 and t=2, and so on up to the between t=4 and t=5

4.) What is the found speed for the intervals.


Homework Equations


I think most of this would be done on Graph mode, on a graphics calculator.


The Attempt at a Solution



I got 125m for the height, 25ms-1 for the overall journey, although i can't remember my answers for the next questions. For the time intervals, i entered Y-cal for the height and entered the distance between each time interval and divided it by the second in between-I'm not sure about this.
Assuming that your distance units are meters and time units are seconds, yes 125 m for the height and 24 m/s for the average velocity are correct. I have no idea what you mean by "Y-cal"- that's not a standard term and there is no mention of it in the problem.

Thankyou for your help. This is very easy, but it's hard to do something that leads you through doing the task incorrectly. Basically in these investigations you have to follow the instructions, and in this case you have to follow the student doing the wrong formulas, etc. Although as the paper unfolds the student finds the correct formula.
It's not clear to me what you are asking! Yes, what you at the end looks correct. You can find f(0), f(1), f(2), f(3), f(4), and f(5). The differences, f(1)- f(0), f(2)- f(1), f(3)- f(2), f(4)- f(3), and f(5)- f(4) give the distance the item fell during that time. Dividing by "1", the time difference gives the average speed during that second.
 
He identifies that his formula ism't correct and so attempts to find the speed at each time interval of t=0 and t=1, then t=1 and t=2, and so on up to the between t=4 and t=5

That doesn't make much sense.

Avg. Speed = |v| = \frac{s_f-s_0}{t_f-t_0}

Which is what you stated, essentially. Don't know why your teacher would say that's wrong for an average speed.

Now using the time intervals might give you a better approximation to the instantaneous speed between the intervals...
 
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