Time taken for body under free fall

Click For Summary

Homework Help Overview

The problem involves a ball dropped from a cliff, with a specific focus on calculating the time taken for its descent after a given time for half of the fall. The context is within the subject area of kinematics, particularly free fall under the influence of gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations to determine the time of descent and height of the cliff. There are questions about sign conventions and the implications of negative time in the calculations. Some participants explore the correct interpretation of distance in relation to the chosen origin.

Discussion Status

The discussion is ongoing, with participants providing guidance on sign conventions and clarifying the equations used. There is an exploration of different interpretations of the problem, particularly regarding the signs of variables in the equations.

Contextual Notes

There is a noted confusion regarding the sign convention for distance and time, as well as the implications of the chosen origin for the coordinate system. The original poster's calculations have led to questions about their validity, particularly concerning negative values.

Swetasuria
Messages
48
Reaction score
0

Homework Statement


A ball is dropped from the top of a cliff. If time taken for half of the descent is 3s, then what is the time taken for the rest of rest of the descent? Take g=10m/s2


Homework Equations


T=√(2h/g)
v=u-gt
s=ut-gt2/2

The Attempt at a Solution


T=√(2h/g)
h=T2g/2=45m

∴height of cliff=90m

v=u-gt=0-(10*3)=-30ms-1

s=ut-gt2/2
45=-30t-5t2
t=-3

:frown: Looks wrong to me. Time can't be negative, can it? Also, shouldn't the answer be less than 3?
 
Physics news on Phys.org
Swetasuria said:
s=ut-gt2/2
45=-30t-5t2
t=-3
You're making a sign error. Use the sign convention consistently. If down is negative, what should be the sign of s?
 
If the origin is on the ground, the distance (which is above the origin) will be positive, right?
 
Swetasuria said:
If the origin is on the ground, the distance (which is above the origin) will be positive, right?
No. What you're calling 's' is really Δs, which is negative as the object is falling.

The full statement of position as a function of time would be (using y instead of s, for clarity):
y = y0 + ut - 1/2gt2

In your case, y = 0 (the final position, measured from the ground) and y0 = 45.

Let me know if that's clear.
 
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
 
Swetasuria said:
Okay. So y=yo+ut-gt2/2
0=45-30t-5t2
t=1.242s
Looks good.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K