Calculating the Height of a Stadium Using Ball Drop: 2.29 Seconds & Speed

AI Thread Summary
A ball dropped from a stadium takes 2.29 seconds to hit the ground, leading to a calculated height of approximately 25.7 meters using the equation h = h_0 - (gt^2)/2. The final speed of the ball upon impact can be determined using the formula v = at, resulting in a speed of about 9.8 m/s after one second. Clarifications were made regarding the difference between acceleration and speed, emphasizing that the acceleration due to gravity is 9.8 m/s², while the speed increases gradually. Additional questions arose about scenarios involving thrown objects, indicating the need for initial height or speed to solve those cases. The discussion concluded with an acknowledgment of the calculations and a request for further assistance on related problems.
thschica
Messages
47
Reaction score
0
A ball is droped from a stadium.It hits the ground 2.29 seconds later.How high is the stadium?Do I use this equation?.5at^2(Thant is wrong isn't it?) How fast is the ball going when it hits the ground?(what equation do I use on this one?)
 
Last edited:
Physics news on Phys.org
I'm assuming no initial speed, then the height h is given by:

h = h_0 - \frac{{gt^2 }}{2}

Here, h_0 is the initial height, so what you are looking for. You choose h = 0, because that's where it hits the ground. Then fill in t and g and solve for h_0.
 
In this case would the answer be about 25.7 meters? And How do I tell How fast the marble is going?
 
That seems to be correct yes.

For the other question, use a relation between acceleration, speed and time. If time is in s and acceleration in m/s², what would give speed (m/s)?
 
would that equation be the y='s one?
 
I was thinking about v = at
 
I have another question.If something is dropped and hits the ground one second later how high is it? With that equation I got 7.1.Why is it not 9.8 meters?
 
Because it's not the speed which is 9.8m/s but the acceleration which is 9.8m/s².
Are you sure you got 7.1 though?
 
No I got 4.9 sorry
 
  • #10
Would the ball be going 22m/s before it hit the ground?
 
  • #11
thschica said:
No I got 4.9 sorry
That is correct.

You see, an acceleration of 9.8m/s² means that after a full second, the speed has increased 9.8m/s. So when dropping something with no initial speed, it only reaches the speed of 9.8m/s after the full second, so when it hits the ground in your example.

The avarage speed was 9.8/2 = 4.9, exactly what you found :smile:

thschica said:
Would the ball be going 22m/s before it hit the ground?
That seems correct, approximately.
 
  • #12
Thank You so much TD
 
  • #13
No problem :smile:
 
  • #14
Say someone threw the ball up and it didnt hit the ground until 3.53 seconds later how do I find out the ending velocity?What if it was thrown down and hit the ground 1.81 seconds later?
 
  • #15
Thrown up would require the initial height and thrown down the initial speed, unless there is none.

Perhaps someone else can help, I'm logging off. 2.50 AM here, good luck!
 
Back
Top