A package is dropped from the plane which is flying with a c

A package is dropped from the 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 at the moment the package is released at A, where it has a horizontal velocity of v_A = 150 ft/s, and just before it strikes the ground at B.

Solution

a.) For the moment it is dropped, the particle will be travelling in a straight direction, hence the radius of curvature would be infinite in that case. Further, there will be no tangetial acceleration as no force is present in the horizontal direction. Whereas, gravitation acceleration will act in the vertical direction and will be equal to 9.81 m/s^2 in magnitude acting downwards, i.e., normal in this case.

b) We know that: y = 1500 - 0.5*g xt^2

and x = 150t

Substituting for t in equation 1 from second equation, we get

Y = 1500 - 0.5*32.2*x^2 / 150*150 is the equation of motion for the particle

Further, 1500 = 0.5*32.2*t^2 ; That is, 9.65 seconds

Vertical velocity at impact = 32.2 * 9.65 = 310.8 ft/s

Hence the angle with the vertical is tangent inverse of 150/310.8 = 25.7 degrees

Hence acceleration in tangential direction = 32.2 cos25.7 = 29 ft/s^2

Acceleration towards the normal = 14 ft/s^2

Further, from the equation of the curve, we have dy/dx = 32.2x/22500

d^2y/dx^2 = 32.2/22500

Radius of curvature for a given curve is obtained as: R = [1 + y\'^2]^3/2 / y\'\' where y\' is single differntiation while y\'\' is double differentiation

Now, we place the values from above to get: R = 7.812 x 10^3 Ft