Constant acceleration simultaneous equation

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A pizza is dropped from a height of 20 m while another is thrown upward at 30 m/s from ground level, and the goal is to determine their meeting point in the air. The acceleration due to gravity is -10 m/s². The equations of motion for both pizzas are set up, with the falling pizza described by x = 0 - 1/2at² and the thrown pizza by x = 30t - 1/2at². The approach involves equating the two equations to solve for time (t) and then substituting back to find the position (x). The discussion highlights the need for simultaneous equations and proper application of kinematic equations to solve the problem accurately.
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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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