A ball is thrown from the roof of a building

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Homework Help Overview

The discussion revolves around a physics problem involving the motion of a ball thrown from a building, described by a displacement function s(t) = 245 + 24.5t - 4.9t². Participants are exploring how to determine the height of the building, the time it takes for the ball to reach the ground, and its velocity upon impact.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of negative time in the context of the problem and question the validity of using t=10 in their calculations. There are attempts to recalculate the position of the ball at t=10 and to clarify the correct approach for finding the velocity function.

Discussion Status

The discussion is ongoing, with participants providing guidance on recalculating values and questioning assumptions about the motion of the ball. Some participants express confusion about fundamental concepts, while others seek clarification on the equations involved.

Contextual Notes

One participant mentions a lack of recent physics knowledge, which may affect their understanding of the problem. There is also a focus on the need to find the correct time when the ball reaches the ground, as well as the velocity function.

plissken170
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It is s(t)=245+24.5t-4.9t(sqrd) meters from the ground (t) after it is thrown. s(t) = displacement function that describes the position of the ball from the ground at time t.

Determine height of building, time it takes the ball to reach the ground and the velocity when it hits the ground.I've been messing around with first, second derivatives and breaking down the original equation into t=10 and t=-5 but I don't know where to go from here.

Help apperciated.

i should mention i took physics 2 years ago and don't remember much of it
 
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Well since the idea of negative time here is worthless, how could you use that t=10?
 
Pengwuino said:
Well since the idea of negative time here is worthless, how could you use that t=10?

Well I plug it in the equation and get 980?
 
What equation did you put t=10 into?
 
s(t)=245+24.5t-4.9t(sqrd)
 
Recalculate that. What do you think the position will be after 10 seconds?

You wouldn't think that after 10 seconds, the object is 980 meters off the ground...
 
Of course I woudn't that's why it doesn't make any sense to me.
I think I am fundementally not getting some kind of simple concept here.
1st derivative when 10 is plugged in i get 1.47
 
Well you were doing the calculation wrong.

245 + 24.5(10)-(4.9)(10)(10)=0 which is the position at t=10.

What equation do you use for finding the velocity in the y direction?
 
v(t)=s dirvative(t)

I don't know. I'm not taking physics. I've covered my calculus books chapter and coudnt find anything.

I'm going to rest on it, thanks for you help though.
 
  • #10
Where is the ball at t=0?

At what time does s(t) = 0 and where is the ball then?

How do you find the function which discribes the velocity as a function of time?
 

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