Projectile motion, 30 and 60 degrees on a level plain

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SUMMARY

The discussion focuses on proving that two projectiles fired at angles of 30 and 60 degrees will travel the same horizontal distance, while the projectile fired at 60 degrees will reach a maximum height three times greater than that of the projectile fired at 30 degrees. Key equations utilized include the kinematic equations for displacement and the conservation of energy principle, specifically relating kinetic energy at launch to potential energy at peak height. The analysis requires a two-dimensional approach to account for both x and y motion, emphasizing the importance of initial velocity and angle in determining the trajectories.

PREREQUISITES
  • Understanding of kinematic equations, specifically for projectile motion.
  • Familiarity with the conservation of energy principle in physics.
  • Knowledge of trigonometric functions, particularly sine, in relation to angles.
  • Ability to analyze motion in two dimensions (x and y directions).
NEXT STEPS
  • Study the derivation of projectile motion equations in detail.
  • Learn how to apply the conservation of energy to different physical scenarios.
  • Explore the effects of varying launch angles on projectile trajectories.
  • Investigate the role of initial velocity in determining maximum height and range.
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of projectile motion concepts.

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Homework Statement



prove that two bullets fired at 30 and 60 degrees will travel the same distance, and the 60 degree shot will travel 3 times as high as the 30 degree.

Homework Equations


S(displacement) Vo(initial velocity) Vf(final Velovity) t(time) a(acceleration due to gravity)

S=volt + .5at^2
S=Vft - .5at^2
S= (Vft - volt)/2
Vf^2= Vo^2 + 2aS
Vf= Vo +at

The Attempt at a Solution



ive tried numerous ways, but i can't seem to get past point of the final velocity at the peak being zero to solve for time in the last equation...and i can't figure out what to do from there
 
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The motion is in two directions; the x- and y-directions. At the peak the speed in the y-direction is zero, but there is still speed in the x-direction. You need to analyse this problem in two-dimensions.
 
Hi

To prove that the 60° shot will travel 3 times as high as the 30°:

You can use the conservation of energy: Kinetic Engergy at the time of the shot equals the potential energy at the peak.

E_{kin}=E_{pot}

For E_{kin} you have to take the angle into account:
E_{kin}=\frac{1}{2}*m*v_{0}^{2}*sin(\alpha)^{2}

Solve for h (=max height) and compare the values for 30 and 60 degrees.

I hope this helps...
 

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