Find Initial Velocity Formula | 30.0m Height

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Homework Help Overview

The discussion revolves around finding the initial velocity of an object thrown vertically that reaches a height of 30.0 m, while neglecting air friction. Participants are exploring the relevant equations of motion and the implications of the given parameters.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various equations of motion and their applicability to the problem, particularly focusing on the equation v² = u² + 2as. There is uncertainty about the initial velocity and the role of acceleration due to gravity.

Discussion Status

The discussion is active, with participants questioning assumptions about the final velocity at maximum height and exploring the relationship between initial and final velocities. Some guidance has been offered regarding the use of specific equations, but no consensus has been reached on the initial velocity calculation.

Contextual Notes

Participants note the absence of time as a given variable, which influences their choice of equations. There is also a focus on understanding the implications of acceleration being negative during the object's ascent and descent.

Brittykitty
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What formula would I use to find the initial velocity?

An object is thrown vertically into the air and reaches a height of 30.0 m. Neglecting air friction, what was the object’s initial velocity?

I don't quite remember how to do this. Is it 30.0m / 9.81[m/s]^2?
 
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There are mainly three equations of motion

[tex]s= ut+\frac{1}{2}at^2[/tex]

[tex]v=u+at[/tex]

[tex]v^2=u^2+2as[/tex]

u= initial velocity
v= final velocity
s= displacement.

Which one has all of the information you need in it? (note: a is acceleration, which is just 'g' in your case)
 
Hello :)
I'm not given the time so I'm assuming its v^2=u^2+2as ?
 
Brittykitty said:
Hello :)
I'm not given the time so I'm assuming its v^2=u^2+2as ?

Yes, also since the maximum height is given you have the final velocity.
 
Thank you! :)
one more question, how would I find the initial velocity? Would it be 0? or 9.81[m/s]^2? don't remember that part either :\
 
Brittykitty said:
Thank you! :)
one more question, how would I find the initial velocity? Would it be 0? or 9.81[m/s]^2? don't remember that part either :\

Remember m/s2 are the units for acceleration, so in your equation, your acceleration is -9.81.

At the maximum height, the object stops rising. So the final velocity is?

When you understand these two things, using the equation you told me it would be v2=u2+2as, you can can find 'u'.
 
rock.freak667 said:
Remember m/s2 are the units for acceleration, so in your equation, your acceleration is -9.81.

At the maximum height, the object stops rising. So the final velocity is?

When you understand these two things, using the equation you told me it would be v2=u2+2as, you can can find 'u'.


the final velocity would be (30.0m) (-9.81)?
 
Brittykitty said:
the final velocity would be (30.0m) (-9.81)?

no, remember the object's goes up, reaches the maximum height and then moves back down to Earth. So during this, the velocity goes from +ve to 'something' (at max) and then -ve as it moves back down to Earth. What is the 'something' ?
 
displacement? :\
 
  • #10
Brittykitty said:
displacement? :\

no no, I am trying to get you to understand what the final velocity is at the maximum height.

Say you throw something up in the air. It does not continuously move upwards, eventually it fall back down, for it to change direction, what must it instantaneously do?


For a similar analogy, say you are driving forwards, for you to move backwards, what should happen to your forward velocity before you start to move backwards?
 
  • #11
It would decrease velocity? =)
 
  • #12
Brittykitty said:
It would decrease velocity? =)

Right yes, as it moves up, it decrease velocity until it reaches what at the max height?

(Because for it to change direction, it needs to stop moving up and then start moving down)
 
  • #13
final velocity?
 
  • #14
Brittykitty said:
final velocity?

If at the maximum height the object instantaneously stops, the final velocity is?
 
  • #15
0? :)
 
  • #16
Brittykitty said:
0? :)

Right! So v=0

v2=u2+2as

v=0
u2+2as = 0


(remember 'a' is negative)
 
  • #17
Thanks! :)
So..
a=-9.81m/s
s=30.0m
v=0
u2??

How would I find the initial velocity now?
 
  • #18
30*-9.81?
 
  • #19
Brittykitty said:
Thanks! :)
So..
a=-9.81m/s
s=30.0m
v=0
u2??

How would I find the initial velocity now?

Brittykitty said:
30*-9.81?

u2-2as=0

rearrange for u2 and then take the square root of both sides.
 

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