Calculating Acceleration Components and Radius of Curvature for Dropped Package

[jfbueno]

I don't really know how to approach this problem but let me give the question first...

A package is dropped from a plane which is flying with a constant horizontal velocity of V_a=150 ft/s. Determine the normal and tangential components of acceleration and the radius of curvature of the path of motion (a) at the moment the package is released at A, where it has a horizontal velocity of V_a=150 ft/s, and (b) just before it strikes the ground at B. They also give you the height of the plan relative to the ground to be 1500 ft.

I'm not entirely sure but wouldn't the tangential component of accelaration be 0 since it is just being released? And the normal component of acceleration is v^2/r however r is another thing you need to solve for in which I am not sure in how to solve.

 

[lzkelley]

Absolutely correct (neglecting air resistance), there is no force acting on the projectile besides gravity, which only acts downward.
v^2/r applies to constant circular motion (we can neglect the curvature of the earth), and not falling packages. The acceleration is due to gravity, what's the equation for that?

 

[tonyh]

As lzkelley says, this is a projectile question - not a circular motion question. Curvature is to do with the rate of bend of a curve, the rate of turning. And the radius of curvature (at a point) is the reciprocal of the curvature (at that point) - it's a calculus problem. So you need the equation of the trajectory, which for your problem is easy to get.

 

[tiny-tim]

Hi jfbueno!

And the normal component of acceleration is v^2/r however r is another thing you need to solve for in which I am not sure in how to solve.

Yes, you're right … but, as you seem to have noticed, r is more difficult to find than the acceleration!

Just use x and y coordinates, and t, and a bit of calculus.

 

[alphysicist]

I don't think you need calculus for this problem. After kinematics is used to find the final vertical velocity, we would know the velocity and acceleration vectors (magnitude and direction) at both initial and final points.

The normal and tangential components of the acceleration are those components perpendicular and parallel to the velocity vector, and we can find their numerical values. Then the normal component is equal to $v^2/r$, where r is the radius of curvature.