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

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Homework Help Overview

The problem involves a dart player throwing a dart horizontally at a speed of 11.8 m/s, with the dart hitting a board 0.31 m below the height from which it was thrown. The discussion centers around determining the horizontal distance from the player to the board, given the vertical drop due to gravity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations to relate vertical motion and time of flight, with some confusion about the interpretation of the vertical distance of 0.31 m.

Discussion Status

Participants are exploring the relationship between vertical and horizontal motion, with some guidance provided regarding the interpretation of the distance fallen. There is an ongoing examination of the calculations related to time and horizontal distance.

Contextual Notes

There is a noted uncertainty regarding the application of the kinematic equations, particularly in determining the time of flight and its implications for horizontal distance. Participants are also clarifying the definitions of the variables involved in the equations.

candyvera
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Homework Statement


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


Homework Equations



V = Vo + at
X = volt + .5at2
v2 = vo2 + 2a(X - Xo)


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.
 
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Hello candyvera, welcome to PF!

candyvera said:

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

Thankfully, you already know what this distance is. It is given in the problem. 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?"
 
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?
 
candyvera said:
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?

Exactly. 0.31 m is the distance traveled in the y direction (i.e. the vertical direction). The dart fell 0.31 m during its flight.
 
cepheid said:
Exactly. 0.31 m is the distance traveled in the y direction (i.e. the vertical direction). The dart fell 0.31 m during its flight.

okay thnx :] are my calculations correct?
.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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