Constant acceleration simultaneous equation

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SUMMARY

The discussion focuses on solving a physics problem involving two pizzas: one dropped from a height of 20 meters and the other thrown upwards at 30 m/s. The acceleration due to gravity is -10 m/s². The key equations used are the kinematic equations for uniformly accelerated motion, specifically x = viT + 1/2aT² for both pizzas. The solution involves setting the equations for the falling pizza and the thrown pizza equal to each other to find the time (t) at which they meet and subsequently calculating their position (x).

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Homework Statement


A pizza is being dropped from a height of 20 m. Another pizza is being thrown up at 30m/s from 0m(the one falling is above the one being thrown). At what position in the air will they meet?

a= -10m/s^2

Homework Equations


Im not sure but i think that some of the b5 kinematics equations would apply. I am pretty sure that this requires a simultaneous equation.


The Attempt at a Solution


first i had x=viT+1/2aT^2

i attempted to get all the t's to one side but got stuck at (2x/vi)/a=(t^2)/vi)+2t/a

i assumed that the t's must be equal. I am unsure whether or not i approached the problem correctly or not
 
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I appreciate you guys for looking at this but i think i got it:
so basically

you have two side

first equation = falling pizza:

x=0-1/2at^2

you set that one, equal to your thrown pizza: x=30-1/2at^2

set them equal to each other, then solve for t then use that t to solve for x
 

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