- #1
Justinjah91
- 3
- 0
- Homework Statement
- You see an enemy archer and decide to position yourself on the side of a hill with a 60 degree grade. The enemy archer is 300 meters from the base of the hill. If the enemy's bow fires arrows at 60 m/s, how high (measured vertically) do you need to be on the hill to be out of range of the enemy archer?
- Relevant Equations
- Kinematics: ##x(t)=x_i+v_it+\frac 1 2 at^2##
Here's a fully typed version of the problem with a diagramMy attempt:
Given the angle of the hill, I know that the horizontal displacement of the arrow and my vertical height on the hill are related by
Now, to find the maximum height, I need to maximize "h" as a function of theta by taking dh/dθ=0. The problem is that I'm having trouble getting "h" as a function of "θ". I'm getting stuck in the algebra somewhere. Anybody have a handy way to simplify this a little further? (I drew it as a cannon on the whiteboard, but same idea)
I feel like there must be a trig substitution or something that I could have employed much earlier in this process to make my life easier.
Given the angle of the hill, I know that the horizontal displacement of the arrow and my vertical height on the hill are related by
##Δx=d+\frac h {Tan(60)}## ...(1)
where d is the distance of the enemy from the base of the hill (300 m). I also know that there is no acceleration in the horizontal direction, and the time of flight of the arrow is the same for the horizontal and vertical directions, leading to the following:##Δx=v_{xi}t~~## ⇒ ##~~t=\frac {Δx} {v_xi}##
##Δy=v_{yi}t-\frac 1 2 gt^2~~## ⇒ ##~~Δy=v_{yi}\left( \frac {Δx} {v_xi} \right) - \frac 1 2 g \left( \frac {Δx} {v_{xi}} \right)^2## ...(2)
If I now substitute eqn (1) into (2), and let ##Δy=h##, I get:##h=v_{yi} \left( \frac {d+ \frac h {Tan(60)}} {v_{xi}}\right)-\frac 1 2 g\left( \frac {d+ \frac h {Tan(60)}} {v_{xi}}\right)^2##
Lastly, I know that the components of my initial velocity are related to the launch angle by:##v_{yi}=v_isinθ##
##v_{xi}=v_icosθ##
Now, to find the maximum height, I need to maximize "h" as a function of theta by taking dh/dθ=0. The problem is that I'm having trouble getting "h" as a function of "θ". I'm getting stuck in the algebra somewhere. Anybody have a handy way to simplify this a little further? (I drew it as a cannon on the whiteboard, but same idea)
I feel like there must be a trig substitution or something that I could have employed much earlier in this process to make my life easier.
Last edited: