Initial velocity of a Ball thrown up

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Homework Help Overview

The discussion revolves around a problem involving the motion of a ball thrown upwards, specifically focusing on determining the initial velocity given a total time of 10.5 seconds until it is caught. The subject area is kinematics, particularly the equations of motion under constant acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore which kinematic equation is appropriate for the problem, with some suggesting the use of the equation s = ut + 1/2 at². Questions arise regarding the variables involved and the reasoning behind choosing specific equations. There is also uncertainty about how to isolate variables and the implications of the given time and acceleration due to gravity.

Discussion Status

The discussion is active, with participants providing guidance on selecting the correct equation and clarifying the relationships between the variables. There is an ongoing exploration of how to manipulate the equations to solve for time and initial velocity, though no consensus or final solution has been reached.

Contextual Notes

Participants note the assumption that the ball is caught at the same height from which it was thrown, and there is a mention of the known values for time and gravitational acceleration. Some participants express uncertainty about the rules for seeking help within the forum.

simpleee
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Can someone one show me a step by step example of how to work out a problem where a ball is thrown up and it is a catched after let's say 10.5s, so how would I find out the initial velocity?
 
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Welcome to PF!

Hi simpleee! Welcome to PF! :wink:

You have s t and a, and you want to find u …

which of the usual constant acceleration equations do you think will help here? :smile:
 
truthfully, I am not quiet sure.
I need help understanding which one to use and why exactly I should use it.
s = ut + 1/2 at squared
 
Oh and thank you for the welcome. =]
 
simpleee said:
truthfully, I am not quiet sure.
I need help understanding which one to use and why exactly I should use it.
s = ut + 1/2 at squared

(try using the X2 tag just above the Reply box :wink:)

ok … there's only three constant acceleration equations, each with four variables, so you just chose the equation with the four variables that you're interested in.

In this case, you have s t and a, and you want to find u, so you choose s = ut + 1/2 at2 because it has all of them. :smile:
 
but i only have 10.5s and 9.8 for gravity.
So how would i know the rest?
 
simpleee said:
but i only have 10.5s and 9.8 for gravity.
So how would i know the rest?

u = 10.5, a = -9.8, and s = 0 (I'm assuming that the person catches it at the same height that (s)he throws it from :wink:).
 
0 = 10.5 t + 1/2 9.8 t2
0 = 10.5 - 4.9 t2
 
simpleee said:
0 = 10.5 t + 1/2 9.8 t2
0 = 10.5 - 4.9 t2

You mean 0 = 10.5 - 4.9 t …

yes, that's right. :smile:​
 
  • #10
Do I get the t by itself now?
I think I might have posted this on the wrong section. .-.
 
  • #11
simpleee said:
Do I get the t by itself now?

Yes … t = 10.5/4.9. :smile:
I think I might have posted this on the wrong section. .-.

That happens quite a lot here! :biggrin:​
 
  • #12
Why did it turn into 10.5/4.9?
Is there any rules on how many times we can ask for help? o.o
 
  • #13
simpleee said:
Why did it turn into 10.5/4.9?

0 = 10.5 t - 1/2 9.8 t2

so 0 = 10.5 - 4.9 t

so 10.5 = 4.9t

so t = 10.5/4.9 :smile:
Is there any rules on how many times we can ask for help? o.o

You have to keep asking until you understand it! :biggrin:
 
  • #14
Good, good.
All the better! ^_^
2.1m/s?
 
  • #15
I guess this problem depends on how strong your esophagus is.

-rlv.zazzle.com-awesome_smiley_photo_sculpture_photosculpture-p153359710604909267220_210.jpg
 

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