Rearrange the Projectile Motion equation

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SUMMARY

The discussion focuses on rearranging the Projectile Motion equation, specifically d = Vit + 1/2at², to solve for time (t). The participants initially struggled to isolate t but were guided towards using the quadratic formula. The final rearranged equation derived is t = (-v ± √(v² + 4(1/2 a)d)) / (2(1/2 a)), which effectively allows for the calculation of time based on distance, initial velocity, and acceleration.

PREREQUISITES
  • Understanding of basic algebra and quadratic equations
  • Familiarity with the Projectile Motion equation
  • Knowledge of the quadratic formula
  • Basic physics concepts related to motion and acceleration
NEXT STEPS
  • Study the derivation and application of the quadratic formula in physics problems
  • Explore advanced topics in kinematics, focusing on projectile motion
  • Learn how to apply initial conditions in motion equations
  • Investigate real-world applications of projectile motion in sports and engineering
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Students studying physics, educators teaching kinematics, and anyone interested in mastering the rearrangement of equations in motion analysis.

mlbmaniaco
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I have a real big problem. My physics teacher has given an extra-credit assignment to my class to figure out. He said we can use whatever we want to figure it out. The problem was to rearrange the Projectile Motion equation (d = Vit + 1/2at2) to solve for t.

So I started out by asking my Math B2 teacher if she knew how to figure it out. After 15 minutes of work she got t = (d/t -vi) 2/a. But she didn't know how to get the second t out of the equation. We both ended up going over to the other math teacher. He got the same thing. Now the three of us are determined to figure out this answer. Can any physics wizard out there help us? All I want to know is how to rearrange the Projectile Motion equation (d = Vit + 1/2at2) to solve for t. For. Example F=ma, to rearrange the equation to sovle for m, it would be m=f/a. So I Want t= ? So can anyone help?
-Puzzled Student
 
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This equation is of the form

[tex] ax^2+bx +c = 0[/tex]

Use the quadratic formula to solve

[tex] x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
 
They'll probably kick themselves (this sounds like the classic case of getting your eyes so close to the problem they can't see it). Surely, both of them have heard of the quadratic formula.

[tex]d=vt + \frac{1}{2} at^2[/tex]
[tex]\frac{1}{2} at^2+vt - d = 0[/tex]
[tex]t = \frac{-v \pm \sqrt{v^2+4 \left( \frac{1}{2} a\right)d}}{2 \left( \frac{1}{2} a \right)}[/tex]
 
Last edited:

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