What is the height of the flagpole in meters?

In summary, the conversation discusses finding the height of a flagpole using the equation d=v0t+1/2at^2. The person asking the question initially used the simplified equation d=(at^2)/2 but was reminded to use the full equation. They then successfully solved for the height of the flagpole by finding the initial velocity and using the concept of average speed.
  • #1
agentsanta
10
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
 
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  • #2
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:
  • #3
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.
 
  • Skeptical
Likes The Duck
  • #4
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.
 
  • #5
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)
 

What is the "Ball Acceleration problem"?

The "Ball Acceleration problem" refers to the calculation of the acceleration of a ball when it is thrown or launched into the air. It involves determining the rate of change of the ball's velocity over time.

Why is the "Ball Acceleration problem" important?

Understanding the acceleration of a ball is crucial in many scientific fields, such as physics and engineering. It allows us to predict the trajectory of a ball and its behavior in different situations, such as when it is thrown or hit by a force.

What factors affect the acceleration of a ball?

The acceleration of a ball can be affected by several factors, including the initial velocity, the angle at which it is thrown, and any external forces acting on the ball, such as gravity or air resistance.

How is the "Ball Acceleration problem" solved?

The "Ball Acceleration problem" can be solved using the laws of motion, specifically Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

What are some real-life applications of the "Ball Acceleration problem"?

The "Ball Acceleration problem" has many practical applications, such as in sports like baseball and football, where players need to accurately throw or kick a ball. It also has applications in industries like aerospace, where understanding the acceleration of objects in flight is crucial.

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