Calculating Range of a Cannon at an Angle

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Homework Help Overview

The problem involves a cannon firing projectiles at a fixed speed and variable angles, specifically asking for the range when fired at an angle of \(\frac{\pi}{6}\) radians. The context includes kinematics and projectile motion, with a focus on calculating horizontal distance while ignoring air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of "range" in this context, with some clarifying that it refers to horizontal distance. Questions arise about the calculation of time of flight and the relationship between angle and range.

Discussion Status

Some participants have provided insights into the calculations involved, including the need to determine the x and y components of velocity and the time of flight. However, there is no explicit consensus on the approach, and multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the maximum range is given as L, leading to discussions about finding a percentage of L for the specific angle of \(\frac{\pi}{6}\). There is an emphasis on showing work to facilitate further assistance.

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Homework Statement


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 [itex]\frac{\pi}{6}[/itex] above the horizontal? Ignore air resistance.

Homework Equations


Four kinematic equations.

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.
 
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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