91Ó°ÊÓ

Travel time A projectile is fired at a speed of 840 \(\mathrm{m} / \mathrm{sec}\) at an angle of \(60^{\circ} .\) How long will it take to get 21 \(\mathrm{km}\) downrange?

Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fixed vector \(\mathbf{C}\) moves in a plane perpendicular to \(\mathbf{C} .\) This, in turn, can be viewed as the following result. Suppose \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) is twice differentiable for all \(t\) in an interval \([a, b],\) that \(\mathbf{r}=0\) when \(t=a\) , and that \(\mathbf{v} \cdot \mathbf{k}=0\) for all \(t\) in \([a, b] .\) Show that \(h(t)=0\) for all \(t\) in \([a, b] .\) (Hint: Start with \(\mathbf{a}=d \mathbf{r} / d t^{2}\) and apply the initial conditions in reverse order.)

Linear drag Derive the equations $$\begin{aligned} x &=\frac{v_{0}}{k}\left(1-e^{-k l}\right) \cos \alpha \\ y &=\frac{v_{0}}{k}\left(1-e^{-k t}\right)(\sin \alpha)+\frac{g}{k^{2}}\left(1-k t-e^{-k t}\right) \end{aligned}$$ by solving the following initial value problem for a vector \(r\) in the plane. $$ \text{Differential equation:}\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}-k \mathbf{v}=-g \mathbf{j}-k \frac{d \mathbf{r}}{d t}$$ $$\text{Initial conditions:}\begin{aligned} \mathbf{r}(0) &=\mathbf{0} \\\\\left.\frac{d \mathbf{r}}{d t}\right|_{t=0} &=\mathbf{v}_{0}=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(\boldsymbol{v}_{0} \sin \alpha\right) \mathbf{j} \end{aligned}$$ The drag coefficient \(k\) is a positive constant representing resistance due to air density, \(v_{0}\) and \(\alpha\) are the projectile's initial speed and launch angle, and \(g\) is the acceleration of gravity.