Calculating horizontal difference in spring-ball situation

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SUMMARY

The discussion focuses on calculating the horizontal distance a steel ball bearing will travel when launched from a compressed spring at a 25° angle. The ball bearing, with a mass of 21 g, achieves a vertical height of 1.33 m, resulting in an exit speed of 5.11 m/s. The user attempts to apply the SUVAT equations and horizontal motion equations, specifically Vx = Vcosθ, but struggles to arrive at the correct horizontal distance of 2.04 m. The correct approach involves breaking down the motion into horizontal and vertical components to accurately calculate displacement.

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foreverconfused
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1.Problem The problem is: calculate the horizontal distance the ball will travel if the same spring is aimed 25° from the horizontal. This is the second part.

The first part was: a small steel ball bearing with a mass of 21 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.33 m. Calculate the speed at which the ball leaves the spring (my correct answer was 5.11 m/s)
2.Relevant equations: Just like the first part I tried using v2=v02+2ax since that still seems relevant. I know I should be incorporating Vx=Vcosθ but I'm not sure how. 3.The attempt at a solution: I set the final velocity to zero and the initial velocity as the answer I had gotten from part one. I set the gravity as acceleration as I had in part one and solved for x (displacement).

0=5.112+2*9.81*x
then x=1.15 m which I know isn't right.

The answer is 2.04 m, I just don't know how to get that.

Also this is my first post so please let me know if I'm missing any information.
 
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You need to break the motion into horizontal and vertical components.
Are you familiar with the SUVAT equations?
What facts do you know about the two components of movement in this problem?
 

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