
#1
Oct2409, 11:09 AM

P: 3

1. The problem statement, all variables and given/known data
A dart player throws a dart horizontally at a speed of 11.8 m/s. The dart hits the board 0.31 m below the height from which it was thrown. How far away is the player from the board? x y vi= 11.8 vi= 0 a=0 a= 9.81 x= unknown y= unknown t= unknown t= unknown 2. Relevant equations V = Vo + at X = Vot + .5at2 v2 = vo2 + 2a(X  Xo) 3. The attempt at a solution i started of with the y part v0t + .5at2 y = 0+.5(9.81)(t2) couldn't solve for the distance since i didn't have the amount of time and i don't know which equation to use after that because all of them involve velocity and i only have the intial velocity of zero. 



#2
Oct2409, 11:17 AM

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P: 5,198

Hello candyvera, welcome to PF!




#3
Oct2409, 02:00 PM

P: 3

You are using this equation to answer the question, "if an object falls 0.31 m while accelerating under gravity, how long does that take?"[/QUOTE]
wait i'm confused, the .31 m is my distance? i thought the question says that the distance is .31m below the height at which it was thrown? 



#4
Oct2409, 02:48 PM

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Solve A Question in 2D Motion A dart player throws a dart horizontally at a spee 



#5
Oct2409, 06:22 PM

P: 3

.31 = (0)t + .5(9.81)t2 t= .251 then you multiply the time twice for the x direction x= (11.8)(.251*2) + .5(0)(.251*2)2 x= (11.8)(.502) x= 5.93 


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