Ball thrown upward and elapsed time

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In summary, a ball is thrown upward from a 100m tall building and it takes 6.5 seconds for the ball to fall back to the ground. The initial speed of the ball can be calculated using the equation x(t) = x0 + v0*t + 0.5*g*t^2, where x0 is the initial displacement (100m), v0 is the initial velocity, t is time (6.5s), and g is the acceleration due to gravity (-9.8m/s^2). Solving for v0, we get an initial velocity of 16.5 m/s. The 0.5 in the equation represents the constant acceleration due to gravity acting on the ball. This
  • #1
dboy83
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A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.


A) The initial speed of the ball
B) Position and speed of the ball after 1 sec and after 4 secs.
C) The maximum height reached by the ball, measured above the ground.

*Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*
 
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  • #2
Hint: initial position.
 
  • #3
so...

radou said:
Hint: initial position.

So, would the initial speed be 0m/s or is that only when the ball is at rest?
 
  • #4
No, the initial speed does not equal zero. Remember, the ball is thrown. You can get the initial speed directly from the equation which represents the displacement of the ball, x(t) = x0 + v0*t + 0.5*g*t^2. Just set the x-axis along the building and everything will be more clear.
 
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  • #5
radou said:
x(t) = x0*t + v0*t + 0.5*g*t^2.
Just x0, not x0*t.
 
  • #6
Thanx, I mistyped. :)
 
  • #7
radou said:
No, the initial speed does not equal zero. Remember, the ball is thrown. You can get the initial speed directly from the equation which represents the displacement of the ball, x(t) = x0*t + v0*t + 0.5*g*t^2. Just set the x-axis along the building and everything will be more clear.


when you say, (t) = x0*t + v0*t + 0.5*g*t^2 does that translate to
initial displacement times time, plus initial velocity times time, plus 0.5 times gravity times time squared?

would the initial displacement equal 100 meters? time 6.5 sec?
so, I just plug and chug...
 
  • #8
Once again, I apologise for my mistype. x(t) = x0 + v0*t + 0.5*g*t^2. Yes, that translates to initial displacement (100 m) plus velocity times time etc. The time is 6.5 seconds. BUT, there is one more condition. The left side of the equation has to equal zero, since at that time the ball fell on the ground, and we have set the coordinate axis x so that the origin (x=0) is at the bottom of the building (the ground), and x = 100 m is the point from which we threw the ball.
 
  • #9
radou said:
Once again, I apologise for my mistype. x(t) = x0 + v0*t + 0.5*g*t^2. Yes, that translates to initial displacement (100 m) plus velocity times time etc. The time is 6.5 seconds. BUT, there is one more condition. The left side of the equation has to equal zero, since at that time the ball fell on the ground, and we have set the coordinate axis x so that the origin (x=0) is at the bottom of the building (the ground), and x = 100 m is the point from which we threw the ball.

so, 0= 100m + v0*6.5s + 0.5*-9.80m/s^2*6.5s^2 ? then I solve for the initial velocity?
 
  • #10
Yes, exactly. I'm glad you noticed g must be negative.
 
  • #11
hmmm

radou said:
Yes, exactly. I'm glad you noticed g must be negative.


ok, I have 0=100m + v0 * 6.5s + 0.5 * -9.80m * 6.5 after I cancel out the s^2. I'm not sure how I add the meters and seconds... shouldn't I have seconds in the demoninator? Would I use pemdas rule to add these?
 
  • #12
dboy83 said:
ok, I have 0=100m + v0 * 6.5s + 0.5 * -9.80m * 6.5 after I cancel out the s^2. I'm not sure how I add the meters and seconds... shouldn't I have seconds in the demoninator? Would I use pemdas rule to add these?

What? You don't have to add any meters nor seconds. You're dealing with numbers only.
 
  • #13
radou said:
What? You don't have to add any meters nor seconds. You're dealing with numbers only.

I have 100 m and 6.5 s. I'm confused now...
 
  • #14
Yes, and you have all you need. You already wrote down the equation you need to solve. Again, 0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2 .
 
  • #15
radou said:
Yes, and you have all you need. You already wrote down the equation you need to solve. Again, 0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2 .

So, how should I go about multiplying and adding the numbers? I tried to multiply and add the numbers out but i got -106.525, which can't be the answer...
 
  • #16
i think i have it

radou said:
Yes, and you have all you need. You already wrote down the equation you need to solve. Again, 0 = 100 + v0*6.5 + 0.5*(-9.80)*6.5^2 .


0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2
= 100 + v0*6.5 + (-207.025)
= -107.025 + v0*6.5
= 107.025= v0*6.5
= 16.5= v0
so, 16.5 m/s =v0 ?
 
  • #17
Yes, looks good. I'm sure you could do this by yourself (the last step), and I don't think such steps need to be discussed here. :smile:
 
  • #18
where did you..

radou said:
Yes, looks good. I'm sure you could do this by yourself (the last step), and I don't think such steps need to be discussed here. :smile:


where did you get the formula from? where does the 0.5 come from in the formula?
 
  • #19
These are basic kinematic equations which can be found everywhere, and it would be weird of your teacher to assign problems to you for which you do not posess the knowledge to solve; I believe he mentioned these equations at some point. :wink:
 
  • #20
thanks

radou said:
These are basic kinematic equations which can be found everywhere, and it would be weird of your teacher to assign problems to you for which you do not posess the knowledge to solve; I believe he mentioned these equations at some point. :wink:

I couldn't find it anywhere in my book. I still don't understand where the 0.5 came from. I really want to know how physics works instead of plugging and chugging. Thanks for your help. Without brainiacs like you, I would be so lost..
 
  • #21
dboy83 said:
I couldn't find it anywhere in my book. I still don't understand where the 0.5 came from. I really want to know how physics works instead of plugging and chugging. Thanks for your help. Without brainiacs like you, I would be so lost..

To be honest, I'm no brainiac, sadly. :wink: Well, the internet is just the right place to start learning physics. I suggest you start googling up things like 'equations of motion', 'basic kinematics', etc.
 
  • #22
formula for b?

dboy83 said:
A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.


A) The initial speed of the ball
B) Position and speed of the ball after 1 sec and after 4 secs.
C) The maximum height reached by the ball, measured above the ground.

*Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*

what formula would I use for B?
 
  • #23
dboy83 said:
what formula would I use for B?

Use the formula we already mentioned and used: x(t) = x0 + v0*t + 0.5*g*t^2. The reason why we 'called' it x(t) is exactly the reason of interest in B) - every time value we 'plug into' this equation will give us the position at that certain moment. So, x(t) means that the position x of the ball depends of the time t. For example, for t = 1 sec, we have x(1) = x0 + v0*1 + 0.5*g*1^2 = something.. (of course, you have to plug in the values of x0, v0, and g, too) I hope you got the point.

For the speed, use the equation v(t) = v0 + g*t.
 
  • #24
I couldn't find it anywhere in my book. I still don't understand where the 0.5 came from.

how good are you at algebra? if you notice in some of the acceleration/displacement formulas, there's a "2" in them. so if you switch the formula around to solve for a different variable, you might end up with a "1/2" = 0.5.

~Amy
 
  • #25
displacment negative or positive?

dboy83 said:
A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.


A) The initial speed of the ball
B) Position and speed of the ball after 1 sec and after 4 secs.
C) The maximum height reached by the ball, measured above the ground.

*Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*

For part B, I found the positions to be 11.6m for 1 sec and -12.4m for 4 sec. I'm not sure if I'm supposed to take into consideration the height of the building and add it to those positions, or just state that those are the positions relative to the release point. I believe the question is asking for the position relative to the release point since part c is more specific and asks for the position relative to the ground. What do you think? Also, for part c, the maximum height is 13.9m relative to the release point but it asks for the height relative to the ground. So, do I add 100m + 13.9m to get the max height relative to the ground? It seems logical to just add the two to get the max height, but the part that confused me is the displacement of the building is -100m not +100m. For some reason, this part confuses me because I thought I was supposed to add the displacement of the building, which is -100m , plus the displacement of the ball from the release point, which is 13.9m, to get the max height...
 
  • #26
displacment negative or positive?

dboy83 said:
A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.


A) The initial speed of the ball
B) Position and speed of the ball after 1 sec and after 4 secs.
C) The maximum height reached by the ball, measured above the ground.

*Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*

For part B, I found the positions to be 11.6m for 1 sec and -12.4m for 4 sec. I'm not sure if I'm supposed to take into consideration the height of the building and add it to those positions, or just state that those are the positions relative to the release point. I believe the question is asking for the position relative to the release point since part c is more specific and asks for the position relative to the ground. What do you think? Also, for part c, the maximum height is 13.9m relative to the release point but it asks for the height relative to the ground. So, do I add 100m + 13.9m to get the max height relative to the ground? It seems logical to just add the two to get the max height, but the part that confused me is the displacement of the building is -100m not +100m. For some reason, this part confuses me because I thought I was supposed to add the displacement of the building, which is -100m , plus the displacement of the ball from the release point, which is 13.9m, to get the max height...
 
  • #27
dboy83 said:
For part B, I found the positions to be 11.6m for 1 sec and -12.4m for 4 sec. I'm not sure if I'm supposed to take into consideration the height of the building and add it to those positions, or just state that those are the positions relative to the release point. I believe the question is asking for the position relative to the release point since part c is more specific and asks for the position relative to the ground. What do you think? Also, for part c, the maximum height is 13.9m relative to the release point but it asks for the height relative to the ground. So, do I add 100m + 13.9m to get the max height relative to the ground? It seems logical to just add the two to get the max height, but the part that confused me is the displacement of the building is -100m not +100m. For some reason, this part confuses me because I thought I was supposed to add the displacement of the building, which is -100m , plus the displacement of the ball from the release point, which is 13.9m, to get the max height...
I don't understand why you are asking the same question in two posts (which is against the rules here) and not responding when someone provides you with an answer. See the post https://www.physicsforums.com/showthread.php?t=132768".

AM
 
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1. What is the initial velocity of a ball thrown upward?

The initial velocity of a ball thrown upward depends on the force applied to throw the ball and the angle at which it is thrown. It can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time elapsed.

2. How does the elapsed time affect the height of a ball thrown upward?

The elapsed time directly affects the height of a ball thrown upward. The ball will continue to rise until it reaches its maximum height, which is when its velocity becomes 0. The longer the elapsed time, the higher the ball will rise before falling back down due to gravity.

3. Why does the elapsed time for a ball thrown upward differ from the elapsed time for it to fall back down?

The elapsed time for a ball thrown upward differs from the elapsed time for it to fall back down because the ball experiences a constant acceleration due to gravity of 9.8 m/s^2 while it is falling. However, while it is rising, the acceleration due to gravity is acting in the opposite direction, causing the ball to decelerate until it reaches its maximum height. This difference in acceleration results in a shorter elapsed time for the ball to fall back down compared to the time it takes to rise.

4. How does air resistance affect the elapsed time of a ball thrown upward?

Air resistance can affect the elapsed time of a ball thrown upward by slowing down the ball's velocity and causing it to take longer to reach its maximum height. This is because air resistance creates a force in the opposite direction of the ball's motion, making it more difficult for the ball to overcome the force of gravity and rise to its maximum height.

5. What is the equation for calculating the elapsed time of a ball thrown upward?

The equation for calculating the elapsed time of a ball thrown upward is t = (v-u)/a, where t is the time elapsed, v is the final velocity, u is the initial velocity, and a is the acceleration due to gravity (9.8 m/s^2). This equation can be used to calculate the time it takes for the ball to reach its maximum height, as well as the total time in the air before it falls back down.

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