Question

Projectile Motion A projectile shot from a gun has weight $\mathit{w}\mathbf{=}\mathit{m}\mathit{g}$and velocity vtangent to its path of motion. Ignoring air resistance and all other forces acting on the projectile except its weight, determine a system of differential equations that describes its path of motion. See Figure 4.9.2. Solve the system. [Hint: Use Newton's second law of motion in the xand ydirections.]

Short answer

$$\begin{gathered} x\left(t\right)=c_1+c_{2}t\text{and} \\ y\left(t\right)=c_3+c_{4}t+\frac{1}{2}g{t}^2 \end{gathered}$$

Step-by-step solution

To Find the differential equation

We have a projectile shot from a gun is being affected by its weight (mg) only, and we have to obtain the differential equation 's system that describes its path of motion as the following technique :

Since there is only one force affects on this shot, then we can apply newton 's law in the horizontal and vertical directions as

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}=0 \\ m\frac{d^{2}y}{d{t}^2}=mg \end{gathered}$$

where x is the horizontal displacement, y is the vertical displacement, and t is the time.

After that, we have to write them in the differential operating style as

$$\begin{gathered} D^{2}x=0 \\ {D}^{2}y=g \end{gathered}$$

For the first differential equation, if we assume that $x=e^{rt}$, then we can have its auxiliary equation as

$r^2=0$

Then we obtain

$$\begin{gathered} x\left(t\right)=c_1{e}^0+c_{2}t{e}^0 \\ =c_1+c_{2}t \end{gathered}$$

is the equation that describes the path of motion horizontally.

For the second differential equation, for the corresponding homogeneous equation $D^{2}y=0$, if we assume that $y=e^{rt}$, then we can have its auxiliary equation as

$r^2=0$

which has the roots

$r_{1,2}=0$

Then we can obtain the homogeneous solution as

$$\begin{gathered} y_h\left(t\right)=c_3{e}^0+c_{4}t{e}^0 \\ =c_3+c_{4}t \end{gathered}$$

After that, also for the second equation we can obtain the particular solution by assuming that $y_p\left(t\right)=A{t}^2$, then differentiate with respect to as

$$\begin{gathered} D{y}_p=2At \\ {D}^2{y}_p=2A \end{gathered}$$

After that, substitute with the second derivative shown in equation (4) into equation (2), then we obtain

$2A=g$

then we have :$A=\frac{1}{2}g$

Then we have the particular solution as

$y_p\left(t\right)=\frac{1}{2}g{t}^2$

Final Answer

Finally from equations (3) and (5), we can obtain the general solution of as

$$\begin{gathered} y\left(t\right)=y_c\left(t\right)+y_p\left(t\right) \\ =c_3+c_{4}t+\frac{1}{2}g{t}^2 \end{gathered}$$

is the equation that describes the path of motion vertically.