Relief Supplies As shown in Figure 5.3.11, a plane flying
Chapter 5, Problem 20E(choose chapter or problem)
Relief Supplies As shown in Figure 5.3.11, a plane flying horizontally at a constant speed \(v_{0}\) drops a relief supply pack to a person on the ground. Assume the origin is the point where the supply pack is released and that the positive x-axis points forward and that positive y-axis points downward. Under the assumption that the horizontal and vertical components of the air resistance are proportional to \((d x / d t)^{2}\) and \((d y / d t)^{2}\), respectively, and if the position of the supply pack is given by \(\mathbf{r}(t)=x(t) \mathbf{i}+y(t) \mathbf{j}\), then its velocity is \(\mathbf{v}(t)=(d x / d t) \mathbf{i}+(d y / d t) \mathbf{j}\). Equating components in the vector form of Newton’s second law of motion,
\(m \frac{d \mathbf{v}}{d t}=m g-k\left[\left(\frac{d x}{d t}\right)^{2} \mathbf{i}+\left(\frac{d y}{d t}\right)^{2} \mathbf{j}\right.\)
gives
\(m \frac{d^{2} x}{d t^{2}}=m g-k\left(\frac{d x}{d t}\right)^{2}, \quad x(0)=0, x^{\prime}(0)=v_{0}\)
\(m \frac{d^{2} y}{d t^{2}}=m g-k\left(\frac{d y}{d t}\right)^{2}, \quad y(0)=0, y^{\prime}(0)=0\)
(a) Solve both of the foregoing initial-value problems by means of the substitutions u=dx/dt, w = dy/dt, and separation of variables. [Hint: See the Remarks at the end of Section 3.2.]
(b) Suppose the plane files at an altitude of 1000 ft and that its constant speed is 300 mi/h. Assume that the constant of proportionality for air resistance is k=0.0053 and that the supply pack weighs 256 lb. Use a root-finding application of a CAS or a graphic calculator to determine the horizontal distance the pack travels, measured from its point of release to the point where it hits the ground.
Text Transcription:
v_0
(dx/dt)^2
(dy/dt)^2
r(t)=x(t)i+y(t)j
v(t)=(dx/dt)i+(dy/dt)j
m dv/dt=mg-k dx/dt^2i+(dy/dt)^2j
m d^2x/dt^2=m g-k(dx/dt)^2,x(0)=0,x^prime(0)=v_0
m d^2y/dt^2=mg-k(dy/dt)^2,y(0)=0,y^prime(0)=0
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