What is the height of the flagpole in meters?

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The discussion revolves around calculating the height of a flagpole based on the time a ball takes to return to the ground after being thrown. The key equation used is d = v0t + (1/2)at^2, where the initial velocity (v0) must be determined. The ball reaches its maximum height at 2 seconds, where its velocity is zero, indicating it was initially thrown with a speed of 20 m/s. The average speed during the ascent is calculated to be 10 m/s, leading to a conclusion that the flagpole's height is approximately 20 meters. The calculations confirm that using 9.8 m/s² for gravity results in a height of about 19.6 meters.
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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
 
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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=VOt +(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.
 
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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)
 
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