Suppose that a mass m is fired by a gun with an angle of $\theta$ with the horizontal and suppose that the initial velocity of the mass is $v_0$ feet per second. Neglect all forces except gravity and air resistance. Air resistance is equal to the $k$ $\times$ velocity of the object in (ft/sec).
1) Take the origin as the position of the gun and x-axis as horizontal and y-axis as vertical. Show that the differential equation of the resulting motion are
$$m\frac{d^2x}{dt^2}+k\frac{dx}{dt}=0$$
$$m\frac{d^2y}{dt^2}+k\frac{dy}{dt}+mg=0 $$
2) Find the solution fo the systems of differential equation in (1)
My work,
I have drawn a diagram and solve for (1). The motion of the object is upwards so air resistance is acting opposite to the motion of the object. Gravity is always downwards.
I am not sure how to do the (2) one.
$$(mD^2+kD)x=0$$
$$x=c_1+c_2e^{\frac{-kt}{m}}$$
$$(mD^2+kD)y=-mg$$
$$y=k_1+k_2e^{\frac{-kt}{m}}-\frac{mgt}{k}$$
Best Answer
$$\begin{array}{rcl} (mD^2+kD)y &=& -mg \\ (mD+k)y &=& C-mgt \\ (D+\frac km)y &=& C'-gt \\ e^{kt/m}(D+\frac km)y &=& C'e^{kt/m}-gte^{kt/m} \\ D(e^{kt/m}y) &=& C'e^{kt/m}-gte^{kt/m} \\ e^{kt/m}y &=& C''e^{kt/m}-\frac mkgte^{kt/m} + \frac{m^2}{k^2}ge^{kt/m} + D \\ y &=& C'' - \frac mkgt + \frac{m^2}{k^2}g + De^{-kt/m} \\ \end{array}$$