- #1
RMalayappan
- 16
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Originally posted in a non-homework forum, so the homework template is missing
I'm following Professor Ramamurti Shankar's video lecture series for Fundamentals of Physics I on Open Yale and I'm hung up on the last question on the second problem set, of which there is no mention on the solution set.
http://oyc.yale.edu/physics/phys-200/lecture-4
The question asks, "Show that if a projectile is shot from a height h with speed v0 the maximum range obtains for for launch angle [tex]\theta = ArcTan(\frac{v_0}{\sqrt(v_0^2+2gh)})[/tex]
I tried this by starting with y=h+v0sinθt-1/2gt2 and x=v0cosθt. I solved for t in the y equation using the quadratic formula and got the final time tf = (v0sinθ+√(2gh+(v0sinθ)2))/g and I substituted it into the x equation and tried differentiating with respect to θ and I got some monstrosity that looks like it won't yield the answer and looks like too much work anyways for the problem. The problem is that everywhere I find this √(2gh+(v0sinθ)2) term, which looks like it should be replaced with a √(2gh+v02) term to get the intended result. Is there an easier approach that I can use to obtain the answer I'm looking for?
http://oyc.yale.edu/physics/phys-200/lecture-4
The question asks, "Show that if a projectile is shot from a height h with speed v0 the maximum range obtains for for launch angle [tex]\theta = ArcTan(\frac{v_0}{\sqrt(v_0^2+2gh)})[/tex]
I tried this by starting with y=h+v0sinθt-1/2gt2 and x=v0cosθt. I solved for t in the y equation using the quadratic formula and got the final time tf = (v0sinθ+√(2gh+(v0sinθ)2))/g and I substituted it into the x equation and tried differentiating with respect to θ and I got some monstrosity that looks like it won't yield the answer and looks like too much work anyways for the problem. The problem is that everywhere I find this √(2gh+(v0sinθ)2) term, which looks like it should be replaced with a √(2gh+v02) term to get the intended result. Is there an easier approach that I can use to obtain the answer I'm looking for?