Question

Need help solving these two second order differential equations, the picture below are the answers to them. I'm just stuck on how to work them out to get those solutions

3 Assignments 3.1 Analysis in the absence of air resistance Ignoring alr resistance, Newton's equations of motion for an object under the influence of gravity alone are quite straightforward. We're also ignoring the fact that the gravitational pull on an object above the earth depends upon its height above the earth. We take g = 9.81 m/s2. a ply(t) = -9, ex(t) = 0. (1) 1. Solve these two second order differential equations (1) for x(t) and y(t) and use MATLAB or Excel to plot the (x,y) trajectory obtained for angle of inclination 9 = 35°. Observe that x(0) and y(0) were determined by the set-up of my measurements and are known to be 0 and. 18 m respectively. The remaining two initial conditions x (0) and y(0) are given by x'(0) = v cos, y'@) = v sin 8, Where v is the initial velocity with which the gun fires its darts. 2. Use the experimental data from Section 2 to determine the initial velocity with which the darts are fired 3. Determine the distance, d), your object travels as a function of e, and compute the angle that maximizes its distance. Your answer should be in degrees and accurate to two decimal places. Plot this function in MATLAB or Excel and compare it with the experimental values. Discuss the discrepancies, 4. Compute an error, E. for your model based on the sum of squared errors. 27 E- (dexperimental - det?

Answer 1

The way to solve these two second order differential equations is very simple. I am solving them in details below.

$\frac{d^2}{dt^2}x(t)=0,\:\Rightarrow\:\frac{d}{dt}x(t)=c_1\text{ [ Integrating one time with respect to \textit{t} ]}$

Now, we have been given one initial condition that $x'(0)=v\cos\theta$ , so, we see that $c_1=v\cos\theta$ . Next, we have to do the integration another time with respect to t, as is done below:

$\int_{x_0}^{x(t)} dx=v\cos\theta\int_0^t dt,\:\Rightarrow\:x(t)-x_0=v\cos\theta.t,\:\Rightarrow\:\boxed{x(t)=x_0+v\cos\theta.t}$ (Answer)

where, $x_0$ is another constant which can be determined from another initial condition, if given.

The next one is:

$\frac{d^2}{dt^2}y(t)=-g,\:\Rightarrow\:\frac{d}{dt}y(t)=-gt+c_2\text{ [ Integrating one time with respect to \textit{t} ]}$

Now, we have been given one initial condition that $y'(0)=v\sin\theta$ , so, we see that $c_2=v\sin\theta$ . Next, we have to do the integration another time with respect to t, as is done below:

$\int_{y_0}^{y(t)} dy=v\sin\theta\int_0^t dt-g\int_0^ttdt,\:\Rightarrow\:y(t)-y_0=v\sin\theta.t-\frac{1}{2}gt^2,\:\Rightarrow\:\boxed{y(t)=y_0+v\sin\theta.t-\frac{1}{2}gt^2}$ (Answer)

where, $y_0$ is another constant which can be determined from another initial condition, if given.

And Done! Cheers!