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

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically the final speed of a steel ball launched from a height. The ball is launched at an angle of 25 degrees with an initial speed of 10 m/s from a platform 15 meters above the ground.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the conservation of energy principle and uses a kinematic equation to find the final speed. Some participants question the accuracy of the calculations and the relevance of the launch angle.

Discussion Status

Participants are engaged in verifying the calculations presented by the original poster. There is acknowledgment of a potential error in the input values used in the formula, but no consensus has been reached on the final speed.

Contextual Notes

The original poster mentions that this problem is for a test, indicating a need for accuracy in their calculations. There is also a focus on ensuring the correct application of physics principles without providing complete solutions.

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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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