How Do You Calculate the Maximum Height of an Object Thrown Upward?

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Homework Help Overview

The discussion revolves around calculating the maximum height of an object thrown vertically upward, given its speed at a specific height. The original poster expresses difficulty in determining the initial velocity, which is not directly provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore reframing the problem to use known values at a fraction of the maximum height. There is discussion about applying kinematic equations, but some participants question the relevance of specific height fractions and the values of variables involved.

Discussion Status

The conversation is active, with participants suggesting different approaches and clarifying the reasoning behind using certain height fractions. There is no explicit consensus on the best method yet, but several productive lines of inquiry are being explored.

Contextual Notes

Participants note the challenge of working with unknown variables and the implications of the problem's setup, particularly regarding the initial velocity and the specific heights referenced in the discussion.

finlejb
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An object is thrown vertically upward such that it has a speed of 21 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

I can't figure what to do since initial velocity isn't given and I don't see a way to solve for it.
 
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Reframe the problem this way: An object is thrown upward with an initial speed of 21 m/s and rises to a height of h/3. Solve for h.
 
Oh I like that suggestion Doc Al. Very clever.

I would use a kinematic equation to help get your answer... this one perhaps?

[tex]v_{f}^2=v_{i}^2+2ad[/tex]
 
That equation doesn't really help me though, Jameson. It has two variables in it that I don't know the value of.

And I kind of see where you're going, Doc Al, but why h/3? It seems like 2h/3 would be what I'd want to solve for...
 
What doc al is saying is you can find the maximum heigth because at the point [itex]\frac{2}{3}h[/itex] the initial velocity would be 21ms/2 and at maximum height (which is [itex]\frac{1}{3}h[/itex] above that point) you know the ball would come to rest. This would allow you to use Jameson's equation to calculate the maximum height.
 
finlejb said:
And I kind of see where you're going, Doc Al, but why h/3? It seems like 2h/3 would be what I'd want to solve for...
Hootenanny already explained it, but let me put it this way. You have three points of interest:
(1) start: y = 0; v = ?
(2) "mid" point: y = 2h/3; v = 21 m/s
(3) top: y = h; v = 0 m/s

What I'm suggesting is that between points 2 & 3 the object travels a distance of h/3. A perfect opportunity to apply Jameson's kinematic equation.
 

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