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

AI Thread Summary
A dart player throws a dart horizontally at 11.8 m/s, which hits a board 0.31 m below the launch height. The vertical motion is analyzed using the equation for distance under constant acceleration due to gravity, confirming that 0.31 m is the vertical distance fallen. The time of flight is calculated to be approximately 0.251 seconds. The horizontal distance to the board is then determined by multiplying the horizontal speed by the total time of flight, resulting in a distance of about 5.93 meters. This solution effectively combines both horizontal and vertical motion principles in 2D kinematics.
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 = Vot + .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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