Rearranging the Projectile Motion Base Quadratic Formula for Initial Velocity

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Homework Help Overview

The discussion revolves around rearranging the projectile motion equation -16t^2 + Vt + h to solve for the initial velocity V. The original poster provides specific values for height and time, expressing uncertainty about their rearrangement process.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods of rearranging the equation to isolate V. Some question the consistency of the original equation with standard physics equations for motion. Others provide alternative rearrangements and express opinions on the efficiency of their approaches.

Discussion Status

The discussion is active, with participants offering various perspectives on the rearrangement process. While some affirm the original poster's approach, others suggest alternative methods without reaching a consensus on which is superior.

Contextual Notes

Participants note the original poster's uncertainty about their answer and the potential inconsistency with standard physics equations. The context includes specific values for height and time, which are integral to the problem.

Pearlhammer
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Homework Statement


What I'm supposed to do is to rearrange the this formula -16t^2+Vt+h and solve it for V.
V= Initial Velocity t= time (throwing the ball in a parabolic arc) h= height
I know what the height is and it is 6ft. I also have the time which is 2.03 seconds.
How do I rearrange for V? I got an answer but I'm not sure and doubting if it is correct.

Homework Equations



-16t^2+Vt+h

-16t^2+Vt+6 (if you plug in the height.)

The Attempt at a Solution



-16t^2+Vt+6 What I started with

-16t^2+Vt= -6 Subtracted 6 to the other side

Vt= -6+16t^2 Added -16t^2 to both sides.

V= (-6/t)+16t Divided both sides (every term) by t. This is my answer so far.

I have no idea if I'm correct or not and I have the feeling I'm not. Please help, I'm in US grade 9.
 
Last edited:
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Is that meant to be the physics equation for motion:

[tex]h=Vt+1/2gt^2[/tex] ?

So then re-arranging you have [tex]-1/2gt^2-Vt+h=0[/tex]

It doesn't seem to be consistent with what you started with. But if we assume you started with the correct equation, yes, you've done it right.
 
Thank you. What I was talking about was just simply if I had the correct answer for:
Taking -16t^2+Vt+h
and rearranging for V to be by itself on one side of the equation.
 
-16t^2+Vt+h = 0
-16t^2+Vt = -h
Vt = -h + 16t^2
V = (-h + 16t^2)/t

your right op, but my way is better
 
Thank you very much guys!
 
ktgster said:
your right op, but my way is better

How is your way better? It's exactly the same...
 

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