Here is the problem:
A boy stands at the peak of a hill which slopes downward uniformly at angle ϕ. At what angle θ from the horizontal should he throw a rock so that it has the greatest range?
I realize that the same question is posted here: https://physics.stackexchange.com/questions/24235/trajectory-of-projectile-thrown-downhill, but I have some questions that were not answered in that thread:
- Can the problem be solved without a rotation of the coordinate system? If so, how?
- I tried to solve the problem using a rotated coordinate system, but cannot figure out how to finish it (see the work given below).
Here is what I have so far:
- We set up the coordinate system so that the positive x axis coincides with the downward slope of the hill. This simplifies the problem by allowing us to easily relate ϕ and θ, through the relation α=ϕ+θ.
- v0x=v0cosα
- v0y=v0sinα
- ax=−gcos(ϕ−π2)=−gcos(−(π2−ϕ))=gcos(π2−ϕ)=gsinϕ
- ay=−gsin(ϕ−π2)=−gsin(−(π2−ϕ))=−gsin(π2−ϕ)=−gcosϕ
- vx=v0x+∫t0ax(t′)dt′=v0cosα+∫t0(gsinϕ)dt′=v0cosα+t(gsinϕ)
- x=x0+∫t0vx(t′)dt′=∫t0(v0cosα+t′(gsinϕ))dt′=t(v0cosα)+12t2(gsinϕ)
- vy=v0y+∫t0ay(t′)dt′=v0sinα+∫t0(−gcosϕ)dt′=v0sinα−t(gcosϕ)
- y=y0+∫t0vy(t′)dt′=∫t0(v0sinα−t′(gcosϕ))dt′=t(v0sinα)−12t2(gcosϕ)
- To find the flight time of the projectile, we find the time at which its trajectory intersects the ground (in this case, the x axis), by setting y=0 and solving for t. y=t(v0sinα)−12t2(gcosϕ)=0v0sinα=12t(gcosϕ)t=2v0sinαgcosϕ
- Substituting t into the equation for x gives us the distance traveled by the projectile as a function of the angles α and ϕ.x=t(v0cosα)+12t2(gsinϕ)x=(2v0sinαgcosϕ)(v0cosα)+12(2v0sinαgcosϕ)2(gsinϕ)x=2v20gcosϕ(sinαcosα)+2v20gcosϕ(sin2αsinϕcosϕ)x=2v20gcosϕ(sinαcosα+sin2αtanϕ)
- I noticed that the solution in the other thread proceeds from here by differentiating x with respect to α, holding ϕ constant, which gives dxdα=2v20gcosϕ(ddα(12(sin(2α)+sin2αtanϕ))dxdα=2v20gcosϕ(cos(2α)+2sinαcosαtanϕ)dxdα=2v20gcosϕ(cos(2α)+sin(2α)tanϕ)This equation allows us to examine how x changes with respect to α. We see that x increases as α increases, up to a certain point, and then decreases as α increases past this value. This means that the graph of x has a relative maximum at the value of α which produces the maximum range.
- We want to find the value of α that results in the maximum range of the projectile. In other words, we must determine the value of α for which the graph of x has a relative maximum. We achieve this by setting dxdα=0=2v20gcosϕ(cos(2α)+sin(2α)tanϕ)Dividing each side by 2v20gcosϕ produces cos(2α)+sin(2α)tanϕ=0Here is where I get lost. It seems like this should be the easy part, because the only thing left to do is solve the above equation for α, but I don't know how to do it. Could someone please explain this part to me?
Additionally, I would like to know whether the problem can be solved without rotating the coordinate system. I originally set out to solve it using the standard rectangular coordinate system, but got bogged down in some equations that seemed to lead nowhere. Thanks for your help.
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