(adsbygoogle = window.adsbygoogle || []).push({}); A particle is fired into the air with an initial velocity of 60m/s at an angle of 54 degrees from the ground. At time t=5.443, what is the radius of curvature of the path travelled by the particle?

I started by coming up with a vector equation for the path travelled by the particle, using the point of launch as the reference point.

[tex]

\overrightarrow r = (60t\cos 54)\widehat{\underline i } + (60t\sin 54 - 0.5g^2 )\widehat{\underline j }

[/tex]

The way i thought to approach this problem was to consider a circle which closely approximates the curve at the given instant.

For an infinitesimal change in time dt, the velocity along the path is given by:

[tex]

v = r\frac{{d\theta }}{{dt}}

[/tex]

I thought that perhaps if i could calculate the speed along the path v and the angular velocity, then i could use those to calculate the radius of curvature at that instant. I am able to calculate the speed easily, but not so sure about the angular velocity.

Am i on the right track, or should i be taking a different approach?

Thanks in advance,

Dan.

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# Radius of Curvature

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