bb1547…: “Due to a noisy home”

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An archer is shooting at a target which is $40$ feet away. Her bow is at a height of $5\unit {ft}$ , and she aims for the bullseye $4\unit {ft}$ above the ground. The bow fires arrows at $160 \unit {ft/s}$ . Assume that the arrow is fired from the point $(0,5)$ and that the target is at the point $(40,4)$ . Find a vector-valued function describing the path of the arrow from the bow to the bullseye of the target. Let the acceleration due to gravity be $-32 \unit {ft}/\unit {s}^2$ .

Start by writing $\vec {b}''(t) = \vector {0,-32}$ .

Antidifferentiate this function.

Let $\uvec {u} = \vector {\cos (\theta ),\sin (\theta )}$ be the direction that the arrow is fired in.

So $\vec {b}'(t) = \vector {0,-32t} + \uvec {u}160$

Antidifferentiate again.

$\vec {b}(t) = \vector {0,-16t^2} + \uvec {u}160t + \vector {0,5}$

Let $t_{\text {target}}$ be the time that the arrow hits the target. So $\vec {b}(t_{\text {target}}) =\vector {40,4}$

Set: $\uvec {u} = \vector {\cos (\theta ),\sin (\theta )}$

Write $\begin{align*} \cos(\theta) 160 t_{\text{target}} &= 40\\ -16t_{\text{target}}^2 +\sin(\theta) 160 t_{\text{target}} +5 &=4 \end{align*}$

Express both $\cos (\theta )$ and $\sin (\theta )$ in terms of $t_{\text {target}}$ , then use the Pythagorean identity: $\cos ^2(\theta ) + \sin ^2(\theta ) = 1$ to solve for $t_{\text {target}}$ .

$t_{\text {target}} = 1/4$

Now express $\cos (\theta )$ and $\sin (\theta )$ in terms of $t_{\text {target}}$ .

You’ll find $\cos (\theta ) = 1$ and $\sin (\theta ) = 0$ .

$\vec {b}(t) = \vector { \answer {160t} \answer {-16t^2+5} }$