Calculating Speed of Ball Leaving Spring Gun

AI Thread Summary
The discussion revolves around calculating the speed of a ball shot from a spring gun, with a spring constant of 28 N/m and a ball mass of 56 g. The spring is initially compressed by 18 cm and releases the ball when compressed by 12 cm. The calculated initial velocity of the ball is approximately 4.24 m/s, and it is noted that horizontal velocity remains constant unless affected by external forces. However, the book's answer is deemed incorrect because it does not address the vertical component of the ball's motion, which is crucial for determining the overall speed when the ball hits the ground. Understanding both horizontal and vertical components is essential for accurate calculations in projectile motion.
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A spring gun (k = 28 N/m) is used to shoot a 56-g ball horizontally. Initially the spring is compressed by 18 cm. The ball loses contact with the spring and leaves the gun when the spring is still compressed by 12 cm. What is the speed of the ball when it hits the ground, 1.4 m below the spring gun?

I calculated Vi = sqrt((28*6^2)/56)) = 4.24264 m/s

Since the horizontal velocity component is independent of the vertical component doesn't the x velocity remain at 4.24 m/s?
 
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Yes x velocity is constant unless acted upon by air resistance or something.
 
007banshee said:
Yes x velocity is constant unless acted upon by air resistance or something.


Then why does the book insist my calculated answer is incorrect?
 
Yes, horizontal velocity remains constant. The problem didn't ask for the horizontal component of the velocity.

(hint)
 
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