What Are the Two Angles to Fire a Shell to Hit a Target 1000m Away?

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SUMMARY

The discussion focuses on calculating the two angles required to fire a shell with a muzzle velocity of 350 m/s to hit a target 1000 m away. Key equations mentioned include the kinematic equations for constant acceleration, specifically v = u + at, s = ut + 1/2at², and v² = u² + 2as. The solution involves separating the horizontal and vertical components of motion and using simultaneous equations to eliminate time (t) to find the launch angle (θ). This method effectively addresses the projectile motion problem.

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Awsom Guy
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Homework Statement


A shell is fired with a muzzle velocity of 350ms/s at a target 1000m away. Calculate two angles at which the shell can be fired .


Homework Equations


I think these equations might help...
v=u+at
s=ut+1/2at^2
v^2=u^2 + 2as
Hope these eqations are right.

The Attempt at a Solution


I thought about maybe this way:
cos(theta)=...
Sorry that is as far as I can go...
Thanks for any help.
 
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Hi Awsom Guy! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)

Yes, those three equations will solve any constant acceleration problem.

The trick in this case is to write two lots of equations, one for the horizontal components of velocity (with a = 0), and one for the vertical (with a = -g).

Call the time taken "t", and the launch angle θ, and get two simultaneous equations from which you can eliminate t to find θ. :smile:
 
Ok I will try...
 

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