Calculate Horizontal Distance for 12.0g Steel Ball, 32 Degrees

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To calculate the horizontal distance a 12.0g steel ball travels when launched at a 32-degree angle, the initial velocity was determined to be 5.24 m/s using the equation v^2 = V0^2 + 2a(x - x0). The range formula R = (V0^2 * sin(2θ)) / g was applied, resulting in R = (5.24^2 * sin(64 degrees)) / 9.8. It was noted that the mass of the ball does not affect the horizontal distance due to the independence of motion from mass, as per Newton's second law. The discussion emphasizes the importance of understanding projectile motion and the role of initial velocity and angle in determining range. The calculations and concepts presented are crucial for accurately predicting the ball's trajectory.
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A small steel ball bearing with a mass of 12.0g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.41m. Calculate the horizontal distance the ball would travel if the same spring were aimed 32 degrees for the horizontal.

I found the initial velocity by pluging it into v^2 = V0^2 + 2a(x -x0) which was 5.24m/s, but I'm at a loss of how to proceed from there.
 
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Does it make sense to you that the ball's initial speed must be V0 in the second case?
 
thanx! yes it does... I ended up using the equation R= (5.24)^2 sin(2(32 degrees)) / 9.8 but I'm still a lil unsure if that is right because it does not use the mass at all.
 
Since the force of gravity is the only force during the motion, and that the said force is proportional to the object's mass, it follows from Newton's 2.law that the motion is independent of the mass
 

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