Answer
Result See result
Work Step by Step
Step 1 In this problem, we are to simply follow the proofs found in the chapter listed. This is a position function for projectile motion at a given angle. Step 2 To start, we are given a couple of variables in the equation, and we shall explain each of these variables before we see how this equation came to be. The first one, \(v_0\), is the initial resultant velocity of the projectile. The second one, \(\alpha\), is the angle between the projectile trajectory and the horizontal. Next is \(t\), which is the point of time in which you want to know the projectile's position. \(s_0\) is the initial position of the shell, which in this case, acts as the y-intercept. Further, \(g\) is simply the acceleration due to gravity. All of these values affect the position of the projectile. Of course, let us not forget about the last ones, which are \(\mathbf{i}\) and \(\mathbf{j}\). As you know, these are unit vectors signifying the particle's position in both the \(x\) and \(y\) directions. Step 3 For the derivation, we are to have several assumptions before solving them. The first one is the mass of the object. It would be assumed that the mass is always constant. Second, there is no acting force on the particle aside from gravity (no drag or other kinds of attractive/repulsive forces). Third, we assume acceleration to be constant. Step 4 From here, we should know that it was derived using Newton's formula and was integrated until position has been reached. To be specific, here are the steps in how it was derived. \[ \mathbf{F} = m\mathbf{a} \] Now, since we are talking about vectors, there should be unit vectors that reside in both \(x\) and \(y\) axes. However, since we assumed that gravity is the only force that affects our motion, we can say that there is only force, and that is found in the y-component. We let this acceleration equal to the gravitational constant, \(-g\). We equate these info with equation one, giving us the following relation. \[ m\mathbf{a} = -mg\mathbf{j} \] Now, we integrate acceleration with respect to time, giving us velocity. \[ \mathbf{v}(\mathbf{t}) = \int -g\mathbf{j}\,dt \] Since mass and gravity are constant, we are to only integrate \(-g\,dt\). This gives us the following answer. \[ \mathbf{v}(\mathbf{t}) = -gt\mathbf{j} + \mathbf{c}_1 \] Setting \(t = 0\) will give us the velocity at \(t = 0\), therefore, \(\mathbf{c}_1 = \mathbf{v}_0\). \[ \mathbf{v}(\mathbf{t}) = -gt\mathbf{j} + \mathbf{v}_0 \] We now have the equation for velocity. Step 5 However, since we are talking about the position, we are to further integrate the velocity equation with respect to time, giving us position. However, in this case, keep in mind that the horizontal position would be kept at a baseline, in which \(x(0) = 0\), \(y(0) = s_0\mathbf{j}\), and when you add them together, \(r(0) = s_0\mathbf{j}\). Integrating the velocity, we get the following expression. \[ \mathbf{r}(\mathbf{t}) = \int -gt\mathbf{j} + \mathbf{v}_0\,dt \] Mass and gravity are constant, so we are to only integrate \(-gt\mathbf{j}\,dt\). This gives us the following answer. \[ \mathbf{r}(\mathbf{t}) = \left(-\frac{1}{2}gt^2\mathbf{j}\right) + \mathbf{v}_0t + \mathbf{c}_2 \] Now, since we have these, let us get \(\mathbf{c}_2\) via our initial condition, \(\mathbf{r}(0) = s_0\mathbf{j}\). To do that, we set \(t = 0\). Thus, our working equation for the position of the particle in projectile motion is the following. \[ \mathbf{r}(\mathbf{t}) = \left(-\frac{1}{2}gt^2 + s_0\right)\mathbf{j} + \mathbf{v}_0t \] This is not done yet, however. The initial velocity of this particle can still have both an \(x\) and a \(y\) component. Therefore, let us let get the parametric equation for the velocity given at an angle \(\alpha\), where \(\alpha\) being the angle between the trajectory and the horizontal. \[ \mathbf{v}_0 = (v_0\cos\alpha)\mathbf{i} + (v_0\sin\alpha)\mathbf{j} \] Lastly, we combine equation 4 and 5 together, yielding us the following. \[ \mathbf{r}(\mathbf{t}) = \left(v_0\cos\alpha\right)t\mathbf{i} - \left(\frac{1}{2}gt^2 - s_0\right)\mathbf{j} + \left(v_0\sin\alpha\right)t\mathbf{j} \] Simplifying, we get the following final derivation. \[ \mathbf{r}(\mathbf{t}) = \left(v_0\cos\alpha\right)t\mathbf{i} - \left(\frac{1}{2}gt^2 - s_0\right)\mathbf{j} + \left(v_0\sin\alpha\right)t\mathbf{j} \] Which is equal to the equation in question.
