Highland Games Hay Toss: Solving for Time and Direction in 2D Kinematics

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The discussion revolves around solving kinematic equations for a hay toss event in the Highland Games. The initial velocity of the hay bale is given, and the user successfully calculated the time for the bale to reach a speed of 7 m/s using the quadratic equation, arriving at 0.201 seconds. However, they struggled with determining the time when the hay bale is moving at a 45-degree angle below the horizontal. A hint was provided to focus on the vertical component of the velocity, suggesting that when the vertical component reaches 5 m/s downward, the problem can be solved more easily. The user ultimately recognized that they were overcomplicating the analysis.
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2D Kinematic Equation! Need Help! I have a test tomorrow!

Homework Statement


One of the most popular events at highland games is the hay toss, where competitors use a pitchfork to throw a bale of hay over a raised bar. Suppose the initial velocity of a bale of hay is varrowbold = (1.32 m/s)xhatbold + (8.85 m/s)yhatbold.
(a) After what minimum time is its speed equal to 7 m/s?

(b) How long after the hay is tossed is it moving in a direction that is 45.0° below the horizontal?





Homework Equations


I figured out the answer to part A using the quadriatic equation by saying that ...
t=2Vnotyg+-sqrt of (-2Vnotyg)^2 - 4(g^2)(Vnotx^2+Vnoty^2-V^2)/2(g^2).
I came up with the right answer of .201s.

I am stuck with part B. I don't know if I am overanalizing it or what. Someone please help!
 
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matt@USA said:

Homework Statement


One of the most popular events at highland games is the hay toss, where competitors use a pitchfork to throw a bale of hay over a raised bar. Suppose the initial velocity of a bale of hay is varrowbold = (1.32 m/s)xhatbold + (8.85 m/s)yhatbold.
(a) After what minimum time is its speed equal to 7 m/s?

(b) How long after the hay is tossed is it moving in a direction that is 45.0° below the horizontal?





Homework Equations


I figured out the answer to part A using the quadriatic equation by saying that ...
t=2Vnotyg+-sqrt of (-2Vnotyg)^2 - 4(g^2)(Vnotx^2+Vnoty^2-V^2)/2(g^2).
I came up with the right answer of .201s.

I am stuck with part B. I don't know if I am overanalizing it or what. Someone please help!

When it is moving at 45 degrees, the verticla (y) and horizontal (x) component are the same size.
At 45 degrees down means the y-component is 1.32 down.
 
So I would say 1.32sin45, which would equal what? I am confused ... My mind is scrambled right now.
 
matt@USA said:
So I would say 1.32sin45, which would equal what? I am confused ... My mind is scrambled right now.

The original velocities given didn't have a sin45 component, so why did you put one in.

Hint: at what time will the hay bale have a vertical component of 5 m/s down? If you can work that out, you can solve part (b) in a similar way
 
So I would use the Vy=Voy+Ayt right? And since Voy=0, the equation would be t=Vy/Ay?
 
Nm, I got it! It was t=x+y/g! I am over analizing this stuff way too much!
 
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