Question

Assume a coordinate system where the origin is the point where supply pack is released, the positive x axis is the direction the plane is flying and the positive y axis points downward.   Assume the horizontal and vertical components of the air resistance are proportional to the square of the velocities

1. Using the appropriate model using accuracy to the nearest foot rounding off at the end of the problem, determine the horizontal distance the pack travels assuming

The plane altitude is 1000 feet above the ground, the plane’s speed is constant at 300 miles per hour, the proportionality constant for air resistance is k=0.0053 in the correct units, and the pack weighs 256 pounds

Answer 1

given coordinate axes, y is downwards, x is to the right, pack is released at origin
air resistance is proportional to square of velocity

so in y direction
my" = mg - kVy^2
and in x direction
mx" = -kVx^2
also, Vy = dy/dt and Vx = dx/dt
hence
my"= mg - ky'^2
mx"= -x'^2

initial conditions, if time of flight is T
y(T) = 1000 ft
Vx(0) = 300 miles per hour = 440 foot per second
mg = 256 lb
and k = 0.0053
hence,

y"= 32 - 0.0053y'^2 = -0.0053(y^2 - 6037.73584)
vy*dvy = -0.0053(y^2 - 6037.73584)dy
integrating
vy^2/2 = -0.0053(y^3/3 - 6037.73584y) + C1
using vy = 0 at y = 0
0 = C1
hence
vy^2/2 = -0.0053(y^3/3 - 6037.73584y)
vy = sqrt(-2*0.0053(y^3/3 - 6037.73584y)) = dy/dt
dt = dy/sqrt((63.999999904y - 0.00353333333333y^3))
integrating from y = 0 to y = 1000 ft we get total time of flight T

for the horizontal motion
mdx'/x'^2 = -dt
integrating
m[1/x'(0) - 1/x'(f)] = -t
x'(f) = m/( m[1/440] + t)
dx = m*dt/( m[1/440] + t)
integrating
x = 256*ln(( 1 + 440t*32/256))/32
so
total horizontal distance travelled is
x(T) = 8*ln(1 + 55*integral(dy/sqrt((63.999999904y - 0.00353333333333y^3))))