How High Should I Drop My Weapon Sights for a 2.2 M/S Recoil Simulation?

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SUMMARY

The discussion focuses on calculating the drop height required for two weapon sights weighing 2.2 lbs and 2.6 lbs to achieve a velocity of 2.2 meters per second upon impact with the ground. Utilizing the principle of conservation of energy, the equation mgh = 1/2 mv² is applied, where g is the acceleration due to gravity (9.81 m/s²). This allows for the determination of the necessary height (h) for each sight based on the desired final velocity.

PREREQUISITES
  • Understanding of basic physics concepts, particularly conservation of energy.
  • Familiarity with gravitational acceleration (9.81 m/s²).
  • Knowledge of the relationship between mass, height, and velocity in free fall.
  • Basic algebra skills for solving equations.
NEXT STEPS
  • Calculate the exact drop height for both weapon sights using the derived formula.
  • Explore the effects of different weights on drop height calculations.
  • Research real-world applications of recoil simulations in firearms.
  • Investigate additional factors affecting projectile motion, such as air resistance.
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Physics students, firearms enthusiasts, engineers involved in weapon design, and anyone interested in the mechanics of projectile motion and recoil simulations.

amdcursed
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I need to do a rifle recoil simulation to some weapon sights. I have two of them. One weighs 2.2 lbs and the other 2.6 lbs. I need them to be going 2.2 meters a second when the hit the ground.

How high do I have to drop each one?
 
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Seems like a strange thing to do to a rifle.

Objects accelerate downward at g (9.81 m/s2), which is the acceleration of gravity at sea level. If an object starts are rest and falls, then one can use conservation of energy - gravitational energy (mgh) becomes kinetic energy (1/2 mv2), or

mgh = 1/2 mv2, which can be simplified to find h as a function of v.
 

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