Find the Trajectory's Launch Angle

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SUMMARY

The discussion focuses on calculating the launch angle for a cannon firing at a hilltop target with a muzzle velocity of 600 ft/s. The horizontal distance to the target is 2640 ft, and the vertical distance is 1320 ft. The initial calculation using the equation 2θ=arcsin([g*R]/V^2) yields an angle of approximately 6.83 degrees, which is incorrect for hitting the hilltop. The correct approach involves using the projectile motion formula h = xtanθ - 0.5gx²/V²cos²θ to account for the elevation of the target.

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  • Understanding of projectile motion equations
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  • Knowledge of basic physics concepts, particularly gravity (g = 32.2 ft/s²)
  • Ability to manipulate algebraic equations to solve for unknowns
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  • Study the derivation and application of the projectile motion formula h = xtanθ - 0.5gx²/V²cos²θ
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  • Explore advanced projectile motion problems involving varying launch angles
  • Investigate numerical methods for solving non-linear equations in physics
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Homework Statement



A cannon fires at a hilltop target with a muzzle velocity of V = 600 ft/s. The
horizontal and vertical distances to the target are L = 0.5 miles and h = 0.25 miles.
Determine the elevation angle at which the cannon should be fired.

L=2640 ft
h=1320 ft


Homework Equations



2θ=arcsin([g*R]/V^2)

The Attempt at a Solution



2θ=asin([32.2*2640]/360000)

2θ=asin(.2361)

θ=6.8282

This would be the correct angle if the cannon ball were to hit the ground, but the ball is landing on the hilltop. What equation should be used to include the quarter mile in the air?
 

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Use the general formula for the projectile.
h = xtanθ - 0.5gx2/v2cos2θ and solve for θ.
 
Thank you!
 

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