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

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 P: 6 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?
 P: 2 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.
 P: 2,982 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.
 Sci Advisor Thanks P: 3,728 A Cannon Fires Projectiles on a Flat Range at Fixed Speed but with Variable Angle Yes, the "range" is the horizontal distance the projectile travels. What angle will produce the maximum range? What is the speed of the projectile if the maximum range is L?

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