## A Cannon Fires Projectiles on a Flat Range at Fixed Speed but with Variable Angle

1. The problem statement, all variables and given/known data
A cannon fires projectiles on a flat range at fixed speed but with variable angle. The maximum range of the cannon is L. What is the range of the cannon when it fires at an angle $\frac{\pi}{6}$ above the horizontal? Ignore air resistance.

2. Relevant equations
Four kinematic equations.

3. The attempt at a solution
I honestly don't know where to start. What exactly does the question mean by the range of the cannon? Is that the horizontal distance the projectile goes?

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 Yes the range will be the horizontal distance. Since projectiles move horizontally at a constant velocity, it is simply calculated by multiplying the x-component of the velocity by the time of flight. x=v(x)*t where v(x) means the x-component of the velocity. v(x) = v*cos(theta) x = the range The trickier part is calculating the time of flight. This is done by finding the y-component of your initial velocity, then recognizing that the final velocity at landing will be the opposite of this value. Knowing that the projectile will accelerate at -9.8 m/s^2 you can then solve for the time of flight and substitute it back into the first equation. v(y-final) = v(y-initial) + a*t v(y-initial) = v*sin(theta) v(y-final) = -v(y-initial) Hope that helps.
 range is how far away the target is. Given the maximum range is L then they are looking for a percentage of L when the angle is pi/6. show some work and people will help.

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