Vertical Projectile Motion: Finding Speed at Maximum Height

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SUMMARY

The discussion focuses on calculating the speed of an object at half its maximum height during vertical projectile motion, specifically using the equations of motion. The maximum height (h max) is established as h max = (v initial^2)/(2g). Participants clarify the use of the final velocity formula vf^2 = vi^2 + 2ad, where d is set to (1/2)h max and acceleration a is -g. The correct application of these formulas leads to determining the final speed at half the maximum height.

PREREQUISITES
  • Understanding of vertical projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of gravitational acceleration (g)
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of kinematic equations in physics
  • Explore the effects of air resistance on projectile motion
  • Learn about energy conservation in vertical motion
  • Investigate real-world applications of projectile motion in sports and engineering
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Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of vertical projectile motion.

jaded18
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First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance. What is the speed of the object at the height of (1/2)h max? i know that h max = (initial v in the y direction / 2g) and i also know that Kinitial +Uinitial =Kfinal +Ufinal or 0.5(mv^2) = 0.25(mv^2) + mg(.5 h max)

please help! you will be blessed, i swear... :)
 
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your formula is wrong.

hmax = (vinitial^2)/(2g)

now use the equation:

vf^2 = vi^2 + 2ad

use d = (1/2)hmax. use a = -g. use vi = vinitial... and that's it. solve for vf.
 
ok i got it. thanks^^
 
Last edited:

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