How Do You Rearrange the Equation x = x₀ + V₀t + ½at² to Solve for V₀?

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Homework Help Overview

The discussion revolves around rearranging the equation of motion, specifically x = x₀ + V₀t + ½at², to solve for the initial velocity V₀. The context includes a scenario involving a ball thrown from a height and its motion under gravity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss algebraic rearrangements to isolate V₀, with one suggesting specific forms of the equation. There is also a consideration of the known variables and the assumption of constant acceleration.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the implications of the variables involved. Some guidance on rearranging the equation has been provided, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants note that the solution for V₀ depends on the known values of other variables, such as time and acceleration, and question whether the assumption of constant acceleration is valid in this context.

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How can I arrange x = x_o + V_o t + (1/2)(a)(t)^2 so that I can solve for V_o algebraically?
 
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To solve for Vo, one would need to x(t) and t, then it would be just a matter of rearranging the terms, or one writes

(x(t) - xo - 1/2 at2)/t = Vo, or

(x(t) - xo)/t - 1/2 at = Vo


So I'm puzzled about the question.

Finding Vo depends on what other variables are known, and applying the appropriate equation of motion. Does one assume that acceleration is constant?
 
This is the original problem: What's the velocity of a ball thrown vertically from a cliff of 95 meters height that strikes the ground in 5 seconds?

You would use the equation I previously listed but I don't know how to solve for it algebraically.

t = 5 s
a = 9.8 m/s^s
x_o = 0 m
x = 95 m
V_o = ?
 
Well consider how far something can fall under freefall in 5 seconds.

If that distance is greater than 95 m, then the ball must be thrown upward to some point, then it falls downward. Then the ball must travel to some height h, in time tup, then fall from height 95 m + h during time 5 s - tup.

See is this reference is helpful.
http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html
 

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