Projectile problem w/ unkown velocity/time

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An athlete's long jump problem involves calculating the take-off speed given a launch angle of 25° and a landing distance of 6.30 m. The relevant equations include components of projectile motion, specifically relating horizontal distance to velocity and time. The solution approach involves breaking down the motion into horizontal and vertical components, using trigonometric relationships to express time in terms of initial velocity. The final calculation leads to a take-off speed of approximately 8.98 m/s. Understanding these principles is crucial for mastering projectile motion in physics.
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Homework Statement


An athlete executing a long jump leaves the ground at a 25° angle and lands 6.30 m away.
What was his take off speed?

Homework Equations


(In all equations, X can be substituted for Y in entire equation)
Vy= Voy + AyT
Vy^2 = Voy + 2AyY
Y=VoyT + (1/2)AyT^2

The Attempt at a Solution


The answer is 8.98 m/s. I can't figure out how to do this problem, and need to understand this for an upcoming final. I have worked several other similar problems, but none without a known velocity and/or time.
Any insight into how to even begin would be greatly appreciated.
 
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Sorry I edited this because I think i made a mistake don't worry I didn't give up on you.

I'm sorry again I keep getting an answer very close and is probably just off because of my rounding. Though it's been a while since I have done projectile motion and I would hate to show you the wrong way with a final coming up.
 
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The x component of velocity is Vo*Cosθ

His total distance would be given by Vo*Cosθ*t

So ...

X = 6.3m = Vo*Cos25*t

Now you still need time so ... look to the skies.

Vo*Sinθ = g*t and t to max height = Vo*Sinθ/g

But total time = 2*t

Now we can solve. t = 2*Vo*Sin25/g = 6.3/Vo*Cos25

Rearranging Vo2*2*Sin25*Cos25 = 6.3*g

Recognizing that 2*Sinθ*Cosθ = Sin2θ

Vo2 = 6.3*g/Sin50
 
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