Solve Kinematics Problem: Stone Thrown from 165m Building, 4s Travel Time

Click For Summary

Homework Help Overview

The problem involves a stone being thrown vertically downward from a height of 165 meters with an initial velocity of 5.0 m/s. The question seeks to determine the distance the stone travels specifically during its fourth second of travel.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations, with some attempting to clarify the specific time interval for which the distance is being calculated. There is confusion regarding whether the question pertains to total distance after 4 seconds or the distance during the fourth second itself.

Discussion Status

Several participants have offered different interpretations of the problem, with some providing calculations and others questioning the original poster's understanding of the question. There is an ongoing exploration of the correct approach to determine the displacement during the specified time interval.

Contextual Notes

Participants note the importance of distinguishing between total distance traveled and displacement during a specific second, indicating a potential misunderstanding of the problem's requirements.

Below
Messages
15
Reaction score
0

Homework Statement



A student throws a stone vertically downward with an initial velocity of 5.0 m/s from the top of a 165 m building. How far does the stone travel during its 4th second of travel?

Homework Equations



d= vit + 1/2at2


The Attempt at a Solution



Vi = -5.0 m/s
t = 4.0 s
a = -9.81 m/s2
d = ?

I tried it in the kinematics equation d= vit + 1/2at2 and the answer is wrong.
 
Physics news on Phys.org
Well, d=volt+.5at^2
so d should be 98.4m. Did you get that?
 
The answer given is 39 m.
 
Wait, I think it was the d for the 4th second, from 3-4s. Because the first second is 0-1s, second second is for 1-2s, etc.

So
Vf=vo+at
V@3=5+9.8*3 and V@4=5+9.8*4
Vf^2=Vo^2+2ad
(V@4)^2=(V@3)^2+2*9.8*d
Find d
 
Last edited:
d = displacement/distance.
 
Yes, i know.
 
Is anyone going to help solve this problem please?
 
I just gave you the answer.
 
As silvashadow has already pointed out, you are not reading the question carefully. It does not ask for the distance after 4 seconds, it asks for the displacement during its 4th second

Displacement during any given time interval=
(Position at final time)-(Position at initial time).

What is the final time? What is the initial?
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
12K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
3
Views
2K