Projectile Motion: Solving for Explosion Co-ordinates

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SUMMARY

The discussion focuses on solving for the explosion coordinates of an artillery shell fired at an initial velocity of 300 m/s at an angle of 55 degrees. The shell explodes after 42 seconds, requiring the separation of x and y motions for accurate calculations. The user correctly identifies the need to apply the equations of motion: X = Vx0 * t and Y = Vy0 * t + 1/2 * ay * t², where the acceleration in the x-direction is zero and the y-acceleration is -9.8 m/s². The use of trigonometric functions to determine the x- and y-components of the initial velocity is essential for solving the problem.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of kinematic equations for motion in two dimensions
  • Ability to apply trigonometric functions to resolve vectors
  • Familiarity with gravitational acceleration (9.8 m/s²)
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  • Learn how to resolve vectors using trigonometric functions in projectile motion
  • Study the kinematic equations for both horizontal and vertical motion
  • Explore the effects of air resistance on projectile motion
  • Investigate advanced projectile motion problems involving varying angles and velocities
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators looking for examples of artillery trajectory calculations.

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Projectile motion...?

Homework Statement



An artillery shell is fired with an initial velocity of 300 m/s at 55 degrees above the horizontal. It explodes on mountainside 42 s after firing. If x is horizontal and y is vertical, find the (x,y) co-ordinates where shell explodes?

Homework Equations





The Attempt at a Solution



I used rf=ri+vvt+1/2 at2

where ri=0
vi=300 m/s
a=-9.8
t=42-0=42s
and theta = 55 degrees ...

I want to know whether my way is correct ? If it is wrong please provide any suggestions... I don't need the full solution...
 
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For this problem, you need to treat the x and y motions separately.

For example, a=-9.8 m/s2 is true for the y-component of motion, but for the x-component a=0.

Also, for the initial velocity you'll need to use some trig to get the x- and y-components, given that the 300 m/s is at 55°.
 


Thank you! I got it.

I used X=Vx0t and Y=Vy0t+1/2ayt2 equations.
 

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