How Far Does an Object Fall in the Third Second of Vertical Motion?

[jefflee0510]

(Neglect air resistance in all cases)
An object falls down from the top of a high overhanging cliff. How far does it fall in the third second of its motion?

I don't know which formula to use. I know that a=9.8ms^-2 and t=3s I think.
Apparently the answer is 24.5m. But I don't know how to work it out. :C

 

[Bystander]

in the third second

When does the third second of the fall begin? t_initial = ?
When does the third second end? t_final = ?

 

[Orodruin]

Hi jefflee0510 and welcome to Physics Forums!

Please post your homework and homework-like questions in the homework forums. You are also required to show some effort when posting in these forums. Simply saying "I have no idea where to begin is not good enough". See our homework guidelines.

 

[BvU]

Hi Jef, welcome to the homework part of PF :)

The idea here is that you use the homework template

1. Homework Statement
2. Homework Equations
3. The Attempt at a Solution​

And the helpers are are obliged to protest if parts are missing. That explains the reception you got from Oro.

As to your statement: I don't know.. -- That can be solved if you pick something that you think is reasonable (e.g. from here) and start to work on the solution by filling in things from 1.) in the template.

In the mean time, Bystander has already helped you on you way !

 

[HalfLostAndSearching]

I can provide a response to your question. In order to determine the distance an object falls in the third second of its vertical motion, we can use the formula d = 1/2 * a * t^2, where d is the distance, a is the acceleration due to gravity (9.8 m/s^2), and t is the time (3 seconds in this case).

Plugging in the values, we get d = 1/2 * 9.8 m/s^2 * (3 s)^2 = 1/2 * 9.8 m/s^2 * 9 s^2 = 44.1 m.

However, this formula gives us the total distance fallen in the first three seconds. To find the distance fallen in the third second specifically, we need to subtract the distance fallen in the first two seconds. This can be calculated using the same formula with t = 2 seconds, giving us d = 1/2 * 9.8 m/s^2 * (2 s)^2 = 19.6 m.

Therefore, the distance fallen in the third second of vertical motion is 44.1 m - 19.6 m = 24.5 m. This is the answer you mentioned, and it is correct. I hope this helps to clarify the calculation process for you.