Kinematic: Problem based on free fall

Click For Summary

Homework Help Overview

The problem involves a body dropped from a tower, covering a distance of '7x' in the last second of its fall, where 'x' is the distance covered in the first second. The discussion centers around kinematic equations related to free fall and the determination of the time taken to reach the ground.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the distances covered in different time intervals and question the initial and final velocities at specific points during the fall. There is exploration of kinematic equations to relate these variables.

Discussion Status

Participants are actively engaging with the problem, raising questions about the velocities at different intervals and the implications of the body's motion just before it hits the ground. Some guidance has been offered regarding the use of kinematic equations, but no consensus has been reached on the final interpretation of the velocities involved.

Contextual Notes

There is a focus on the assumptions regarding the body's motion just before impact and the consideration of forces acting on the body after it hits the ground. Participants are also discussing the consistency of gravitational acceleration values used in calculations.

NoahCygnus
Messages
96
Reaction score
2

Homework Statement


"A body dropped from the top of a tower covers a distance '7x' in the last second of its journey where 'x' is the distance covered in the first second. How much time does it take to reach the ground ?"

Homework Equations


## S = ut - \frac{1}{2}gt^2##

The Attempt at a Solution


Displacement in the first second;
## -x = 0 - \frac{1}{2}gt^2##
##x = 4.9 m##
Displacement during last second;
## 7x \longrightarrow 7(4.9) \approx 35 m##

That's all I could do. I don't know the distance covered between first second and last second , so I don't know how to utilise the current information I've to find the time .
 
Physics news on Phys.org
NoahCygnus said:
I don't know the distance covered between first second and last second ,
What is the velocity of the body at the start of this interval? What is the velocity of the body at the end of this interval?
 
  • Like
Likes   Reactions: NoahCygnus
cnh1995 said:
What is the velocity of the body at the start of this interval? What is the velocity of the body at the end of this interval?
Initially , it is 0 as the body is dropped from rest.
In the last second , I've no idea what the initial velocity is. Considering that the body hits the ground in the last second , it should come to a stop , doesn't that mean the final velocity should be 0? I'm not sure about it. If that's the case should i be able to find the initial velocity of the last second using the equation ##v^2 - u^2 = 2as## ?
 
NoahCygnus said:
it should come to a stop , doesn't that mean the final velocity should be 0?
No.
cnh1995 said:
What is the velocity of the body at the start of this interval? What is the velocity of the body at the end of this interval?
In other words, what is the velocity of the body at t=1s? What is the velocity of the body 1 second "before" it hits the ground?

You need to use relevant kinematic equations involving time.
 
cnh1995 said:
No.

In other words, what is the velocity of the body at t=1s? What is the velocity of the body 1 second "before" it hits the ground?

You need to use relevant kinematic equations involving time.
At t= 1,
## v = u - gt## as u = 0 , t = 1 , it should be ;
## v= 9.8 m/s##

1 second before it hits the ground , I don't know how to get that. Help me out Sheldon.
 
NoahCygnus said:
1 second before it hits the ground , I don't know how to get that.
You know the displacement and time in this interval. Which equation contains these two terms with unknown initial velocity?
 
cnh1995 said:
You know the displacement and time in this interval. Which equation contains these two terms with unknown initial velocity?
So the time interval is 1 second, using the equation
## -35 = u(1) - 5(1) ##
## u = -30 m/s##
 
NoahCygnus said:
So the time interval is 1 second, using the equation
## -35 = u(1) - 5(1) ##
## u = -30 m/s##
Yes. 30m/s downwards.

Now you have initial and final velocities in that interval. You can calculate the time from that.
 
  • Like
Likes   Reactions: NoahCygnus
cnh1995 said:
Yes. 30m/s downwards.

Now you have initial and final velocities in that interval. You can calculate the time from that.
Final velocity should be
## v = -30 - 9.8##
## v \approx -40##
Now to find the total time.
## -40 = 0 -9.8t##
## t = 4 secs ##
Right ?
 
  • #10
NoahCygnus said:
Final velocity should be
## v = -30 - 9.8##
## v \approx -40##
Now to find the total time.
## -40 = 0 -9.8t##
## t = 4 secs ##
Right ?
Right.
(Be consistent with the value of g. Use either 9.8m/s2 or 10m/s2).
 
  • #11
cnh1995 said:
Right.
(Be consistent with the value of g. Use either 9.8m/s2 or 10m/s2).
Thank you . And you said the final velocity won't be 0. I don't get it , does the body hit the ground at the last second. if so shouldn't the body come to a stop and final velocity be 0? Or you don't take into account the contact force ground will exert and don't even consider its there ?
 
  • #12
NoahCygnus said:
Or you don't take into account the contact force ground will exert and don't even consider its there ?
The body hits the ground with some non-zero velocity.
Its velocity may become zero "after" it hits the ground (or it might bounce off) and that gets you into elastic-inelastic collisions. But whatever happens "after" the collision is a different issue. You only need to consider the time between the instant at which the ball is dropped and the instant at which it hits the ground. Only this interval describes the free fall of the body.
 
  • Like
Likes   Reactions: NoahCygnus

Similar threads

Replies
12
Views
5K
Replies
34
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
40
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
Replies
7
Views
8K
Replies
7
Views
3K