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anonymous
 4 years ago
Problem #9 in problem set II (link below). To find the formula for the range, I set y=0 and got the time for it, then plugged it in the x equation. But then I get an expression for the range and I can't guess which value of the angle maximises it. I tried taking the derivative of R w.r.t. theta and equating to zero but it just gets too complicated.. Any hints please?
anonymous
 4 years ago
Problem #9 in problem set II (link below). To find the formula for the range, I set y=0 and got the time for it, then plugged it in the x equation. But then I get an expression for the range and I can't guess which value of the angle maximises it. I tried taking the derivative of R w.r.t. theta and equating to zero but it just gets too complicated.. Any hints please?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here are the equations: Total time (at y=0) \[t^* =\frac{v_{0}\sin \theta \pm \sqrt{v^2_{0}\sin^2 \theta+2gh}}{g} \] and I chose the + sign of course. x equation: \[x=v_{0}t \cos \theta\] Range: \[R=x(t^*)=\frac{v^2_0 \sin 2 \theta}{2g}\left( 1+\sqrt{1+\frac{2gh}{v^2_0 \sin^2 \theta}} \right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry I forgot the link: http://oyc.yale.edu/physics/fundamentalsofphysics/content/resources/problem_set_2.pdf

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is no 9# problem do u want to derive formula for range?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm sorry problem is in the pdf in the downloads, I don't know why isn't it in the online pdf. Anyway, the problem is: Show that if a projectile is shot from a height h with speed v0, the maximum range obtains for launch angle is \[\theta = ArcTan \left( \frac{v_0}{\sqrt{2gh+v_0^2}} \right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1329034287789:dw so distance=usin@*t(no acceleration in horizontal direction) now wat is the total time taken for projectle i will give u a hint: time takento reach maximum ht=2*time taken to reach top to find time taken to reach top, we have usin@ initaial velocity;v=0 at top acc=g=9.8; vu/at i have given it all i will appreciate it if u work it out here and show me the result

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You should use the nonparametric form of parabollic movement. \[y=x*TAN(\phi)(g/2)*(1+TAN ^{2}(\phi))*(x/V _{0})^{2}\] Here: y=h, and \[\phi=\theta\] \[h=x*TAN(\theta)(g/2)*(1+TAN ^{2}(\theta))*(x/V _{0})^{2}\] \[h=x*TAN(\theta)+(g/2)*(1+TAN ^{2}(\theta))*(x/V _{0})^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then, we must look for a maximum value for x, we get this value after calculate the derivative of x relative to theta and equaling this to zero.\[dx/d \theta=0\] Remember chain rule of derivatives, then you shoul obtain an expression of x in function of theta. replace this value of x in the last expression of my previous post and you will get the answer.
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