91Ó°ÊÓ

Given:

Inclination of slope : $15°$

Arrow is above the horizontal by : $20°$

Speed of arrow : $50m/s$

Height above the ground :$1.75m$

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 $$\begin{gathered} \mathrm{θ}=20°+15° \\ =\mathbf{35}\mathbf{°} \end{gathered}$$.

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

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.

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

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

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