Question

An archer standing on a $15°$ slope shoots an arrow $20°$ above the horizontal, as shown in FIGURE CP4.82. How far down the slope does the arrow hit if it is shot with a speed of $50m/s\mathrm{from}1.75m$above the ground?

Short answer

The arrow hit $\mathbf{}\mathbf{298}\mathbf{.}\mathbf{14}\mathit{m}$ down the slope .

Step-by-step solution

Given information

Given:

Inclination of slope : $15°$

Arrow is above the horizontal by : $20°$

Speed of arrow : $50m/s$

Height above the ground :$1.75m$

Simplifying the problem of the projectile 

We can simplify the problem of a projectile on an inclined plane by rotating the plane to represent the new horizontal and vertical axes.

Let's call the original horizontal and vertical axes $x\mathrm{and}y$, respectively.

We now rotate the axis to match the inclined plane, rotating the previous axis $15°$counterclockwise, and naming the new horizontal and vertical axes $x\text{'}\mathrm{and}y\text{'}axes$.

The archer released the arrow at a $20-$degree angle with regard to the $x-$axis; after rotating the horizontal axis by around $15$degrees, the new initial angle of release will be $$\begin{gathered} \theta =20°+15° \\ =\mathbf{35}\mathbf{°} \end{gathered}$$.

The angle with respect to the $a\text{'}-$axis is $\theta =35°$.

: Calculating the time of the projectile 

We can now determine the time of the projectile using its vertical components thanks to the new angle of release in relation to the new horizontal axis, the $x\text{'}-$axis.

We know the archer was$1.75m$above the ground when the arrow was launched and it landed on the ground; however, because we rotate the horizontal and vertical axes, we can argue that the arrow travelled $1.75m$below the height it was released when it hit the ground.

$$\begin{gathered} ∆y=v_{y}t-\frac{1}{2}{a}_{y}t² \\ \Rightarrow 1.75m=50\frac{m}{s}\sin (35°)\left(t\right)-\frac{1}{2}(9.8\frac{m}{s^2}\cos (15°\left)\right)(t²) \\ \Rightarrow -1.75m=28.68t\frac{m}{s}-4.73·t²\frac{m}{s^2} \\ \Rightarrow 4.73·t²\frac{m}{s^2}-28.68t\frac{m}{s}-1.75m=0 \\ \Rightarrow \underset{disregardingtheunitofmeasurement}{\underbrace{4.73·t²-28.68t-1.75}}=0 \end{gathered}$$

: Calculating how far down the slope does the arrow hits 

Using the quadratic equation$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

we can now solve for$t.$

We'll end up with two values:$t=6.12s\mathrm{and}t=-0.06s$, ignoring the negative time.

Now we can figure out how far down the slope the arrow lands.

role="math" localid="1650775900003" $$\begin{gathered} ∆x=v_{x}t+\frac{1}{2}{a}_{x}t² \\ =\left(50\right)\frac{m}{s}(\cos 35°)(6.12s)+\frac{1}{2}(9.8\frac{m}{s^2})(\sin (15°\left)\right){(6.12s)}^2 \\ =\boxed{\mathbf{}\mathbf{298}\mathbf{.}\mathbf{14}\mathit{m}} \end{gathered}$$