What Is the Final Speed of a Steel Ball Launched from a Platform?

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A steel ball is launched at a 25-degree angle from a 15-meter high platform with an initial speed of 10 m/sec. To calculate the final speed upon impact, the formula Vf² = Vi² - 2g(y - yi) is applied, where g is the acceleration due to gravity (9.8 m/sec²). The user initially calculated the final speed as 13.56 m/sec but made an error by inputting -110 instead of 100 in the equation. The correct approach confirms that the angle of launch does not affect the final speed due to conservation of energy principles. The discussion emphasizes the importance of careful calculations in physics problems.
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A steel ball is launched at 25 degrees above horizontal from the top of a platform that is 15 meters above ground. If the initial speed is 10m/sec, what is the final speed when it strikes the ground?
 
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hi rqu1ntana! welcome to pf! :wink:

show us what you've tried and where you're stuck, and then we'll know how to help! :smile:

(use the standard constant acceleration equations, in the x and y directions separately)
 
The problem is that its for a test and i just want to be sure what i did was right. This is what i did,

I used the formula Vf2=Vi2-2g(y-yi), where Vi2 is the initial velocity, g is gravity=9.8, y is 0, and yi is the height.

from all the data given this is what it looks like, Vf=(102-(2*9.8*(-15)))1/2

The result i get from this is 13.56m/sec, i this right?
 
hi rqu1ntana! :smile:
rqu1ntana said:
from all the data given this is what it looks like, Vf=(102-(2*9.8*(-15)))1/2

The result i get from this is 13.56m/sec, i this right?

yes that looks fine :smile: … you've used conservation of energy (so the 25° doesn't matter)

… except that you keyed in -110 instead of 100 ! :redface:
 

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