What is the height of the flagpole in meters?

  • Thread starter Thread starter agentsanta
  • Start date Start date
  • Tags Tags
    Acceleration Ball
Click For Summary

Homework Help Overview

The problem involves determining the height of a flagpole based on the motion of a ball thrown from the ground, returning after 4.0 seconds, with gravitational acceleration provided as 9.8 m/s².

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations, questioning the initial conditions and the correct application of formulas. There is exploration of the ball's velocity at its maximum height and the implications of using different signs for gravitational acceleration.

Discussion Status

Some participants have offered guidance on the correct equations to use and have pointed out potential misunderstandings regarding initial velocity and the direction of acceleration. Multiple interpretations of the problem and approaches to the solution are being explored.

Contextual Notes

There is a mention of the need to conceptualize the problem without relying solely on formulas, as well as the challenge of understanding the motion of the ball in relation to time and height.

agentsanta
Messages
10
Reaction score
0

Homework Statement


A ball is thrown from the ground to the top of the school flagpole. If it returns to the ground after 4.0 sec, what is the height of the flagpole in meters?
g=9.8m/s^2

Homework Equations


Not really sure.
Tried d= (at^2)/2
Didn't really work out though...

The Attempt at a Solution


d= (at^2)/2
d= ((-9.8m/s^2) * (2s)^2)/2
d= -19.6m

>.> it's negative?

Yeah, everybody on this forum is posting something complicated, but I'm posting something to do with acceleration T_T
I would ask my dad, but he's going to be like "do it yourself" or something
 
Physics news on Phys.org
You are on the right track, however the issue is with your equation. The full equation is d=v0t+1/2at^2

The equation you listed is only for when the initial velocity of the ball is zero. However, if you think through the motion of the ball, there is a point in its path where the velocity is zero.

Also, if you are using g as positive 9.81, when that means is you declared down to be positive and up to be negative. If you make g negative you will get a positive number, but it doesn't really matter either way.

Think about where v will be zero and if you get stuck I'll give you another hint.
 
Last edited:
OK, I think I know what you are saying.
The ball's velocity is 0 at its maximum height, where it cannot get any higher.

So I tried the full equation
d=volt +(at^2)/2
d=VO(2s) + ((-9.8m/s^2) * (2s)^2)/2
d=VO(2s) + ((-9.8m/s^2) * 2s^2)
d=VO(2s) + (-19.6m)

at= Vf-VO
-9.8m/s^2 * 2s= 0-VO
-19.6m/s=-VO
19.6m/s=VO

d= 19.6m/s * 2s - 19.6m
d= 19.6m

I think it's right. Is it? Sorry imma go to bed now. Gotta get to school early.
 
  • Skeptical
Likes   Reactions: The Duck
I think it's right. Is it? Sorry imma go to bed now. Gotta get to school early.

Hopefully you check before you go to school tomorrow because that is perfect.
 
Try to do it in your head without formulas. Then you know that you conceptually get it.

If it takes 4 seconds to return to the ground, then it must have reached max height in 2 seconds.

Since it slows 10 m/s every second, then it must have initially been going 20 m/s.

Initially its speed was 20, at the top it was 0. So its average speed is 10 m/s, and it did this for 2 seconds.

Therefore the flag pole is 20 m high. (or 19.6 if you use 9.8 instead of 10)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
Replies
12
Views
1K
Replies
7
Views
3K
Replies
2
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
6
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K