- #1

Argonaut

- 23

- 17

- Homework Statement:
- An archer on ground that has a constant upward slope of 30.0° aims at a target 60.0 m farther up the incline. The arrow in the bow and the bull's-eye at the center of the target are each 1.50 m above the ground. The initial velocity of the arrow just after it leaves the bow has magnitude 32.0 m/s. At what angle above the horizontal should the archer aim to hit the bull's eye? If there are two such angles, calculate the smaller of the two. You might have to solve the equation for the angle by iteration that is, by trial and error. How does the angle compare to that required when the ground is level, with 0 slope? [Sears and Zemansky's University Physics. 13th edition. Exercise 3.87]

- Relevant Equations:
- Equations of 2D position, velocity, and acceleration with constant acceleration

I worked myself into a trigonometry rut. I've tried two approaches, first by not changing the frame of reference, and second by taking the incline as the horizontal x axis. Here is my second attempt:

Take the incline as the horizontal. Then the coordinates of target T are:

$$

\begin{align}

x_T &= 60.0~\rm{m} \\

y_T &=0

\end{align}

$$

The magnitude of the initial velocity of arrow A and the components of its acceleration are:

$$

\begin{align}

v_0 &= 32.0~\rm{m/s} \\

a_x &= -g*\sin{30.0^{\circ}} \\

a_y &= -g*\cos{30.0^{\circ}}

\end{align}

$$

Let ##\alpha## be the angle of the initial velocity of the arrow above the incline. Then the components of the initial velocity are:

$$

\begin{align}

v_{0x}&=v_0\cos{\alpha} \\

v_{0y}&=v_0\sin{\alpha}

\end{align}

$$

The x and y coordinates of the position of the arrow as a function of time are:

$$

\begin{align}

x_A &= v_{0x}t + \frac{1}{2}a_xt^2\\

y_A &= v_{0y}t + \frac{1}{2}a_yt^2

\end{align}

$$

At time t when the arrow hits the target, ##x_A = x_T## and ##y_A = y_T##. Which means that I have

**two unknown values ##\alpha## and ## t##.**

Since ##y_A = y_T = 0##, I use it to express t as a function of ##\alpha##.

$$

\begin{align}

0 &= v_0\sin{\alpha}~t+\frac{1}{2}a_yt^2 \\

t&=-\frac{2v_0\sin{\alpha}}{a_y}

\end{align}

$$

Then substituting t in the ##x_A## equation, and doing some algebraic manipulation I get:

$$

\begin{align}

x_T &= \sin^2{\alpha}~\frac{2v_0^2a_x}{a_y^2}-\sin{\alpha}\cos{\alpha}~\frac{2v_0^2}{a_y}

\end{align}

$$

For the sake of clarity, let constants ##A=\frac{2v_0^2}{a_y}## and ##B={a_x}/{a_y}##. Then I have the following equation, where ##\alpha## is the only unknown:

$$

\begin{align}

\frac{x_T}{A}&=B\sin^2{\alpha} - \sin{\alpha}\cos{\alpha} \\

&=\sin{\alpha}(B\sin{\alpha} - \cos{\alpha})

\end{align}

$$

And this is where I'm stuck. I've tried various trigonometric identities, double and half-angle formulas, but I couldn't make them work.

(edit: LaTex formatting mistakes)

Last edited: