Finding t in a Projectile Motion Equation

  • Context: High School 
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Discussion Overview

The discussion revolves around solving for the variable t in the projectile motion equation h = vt - 1/2gt^2. Participants explore methods to isolate t and address the nature of the equation as a quadratic form, considering the implications of its solutions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks guidance on isolating t in the equation h = vt - 1/2gt^2, given values for h, v, and g.
  • Another participant suggests recalling Algebra II concepts, specifically the method for solving quadratic equations.
  • A participant emphasizes the quadratic nature of the equation and notes that it can yield two solutions, with a focus on the physically relevant positive solution for t.
  • One participant provides a derived expression for t, t = (v - sqrt(v^2 - 2gh)) / g, based on specific values for h, g, and v.
  • A later reply corrects a notation error regarding the representation of v squared, indicating v^2 instead of v2.

Areas of Agreement / Disagreement

Participants generally agree on the quadratic nature of the equation and the approach to isolate t, but there is no consensus on the specific steps or methods to achieve this, as well as the notation used.

Contextual Notes

There are unresolved issues regarding the manipulation of the equation and the assumptions made about the values of h, v, and g. The discussion does not clarify all mathematical steps involved in isolating t.

Who May Find This Useful

Readers interested in projectile motion, quadratic equations, or those seeking assistance with algebraic manipulation in physics problems may find this discussion relevant.

dooner
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I am trying to work out the problem for t in the equation. h=vt-1/2gt^2. I have the given for h,v, and g. How do I get everything except t on one side of the equation to solve for t.
 
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You learned this in Algebra II. (Substitute x for t).
 
Think back to algebra, that is where you should have learned to solve for "t" when you have "t" to the second power and first power. The buzz word for what you have there is a "quadratic equation"
 
dooner said:
I am trying to work out the problem for t in the equation. h=vt-1/2gt^2. I have the given for h,v, and g. How do I get everything except t on one side of the equation to solve for t.
Note that one has terms of 1, t, t2, which tells one that this is a quadratic equation, so one can solve for the roots of the quadratic equation.

Remember, for ax2 + bx + c = 0, there are two unqiue solutions. However, in physics, one would be the proper solution, for example, the one for which t is positive.

There would be a lot of manipulation to get t or x on one side, and the answers on the other side, but it is doable.
 
I ended up with t=(v-sqrt(v2-2*g*h)/g to get the answer I needed for the given info h=100,g=9.81, and v=50. Thanks
 
wrong v2 is actually v^2
 

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