How to find Final Speed.

  • Thread starter rqu1ntana
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  • #1
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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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  • #2
tiny-tim
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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)
 
  • #3
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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?
 
  • #4
tiny-tim
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hi rqu1ntana! :smile:
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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