## 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?

1. anonymous

Here 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: $R=x(t^*)=\frac{v^2_0 \sin 2 \theta}{2g}\left( 1+\sqrt{1+\frac{2gh}{v^2_0 \sin^2 \theta}} \right)$

2. anonymous

Sorry I forgot the link: http://oyc.yale.edu/physics/fundamentals-of-physics/content/resources/problem_set_2.pdf

3. anonymous

there is no 9# problem do u want to derive formula for range?

4. anonymous

I'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 $\theta = ArcTan \left( \frac{v_0}{\sqrt{2gh+v_0^2}} \right)$

5. anonymous

|dw: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; v-u/a-t i have given it all i will appreciate it if u work it out here and show me the result

6. anonymous

You should use the non-parametric form of parabollic movement. $y=x*TAN(\phi)-(g/2)*(1+TAN ^{2}(\phi))*(x/V _{0})^{2}$ Here: y=-h, and $\phi=-\theta$ $-h=-x*TAN(\theta)-(g/2)*(1+TAN ^{2}(\theta))*(x/V _{0})^{2}$ $h=x*TAN(\theta)+(g/2)*(1+TAN ^{2}(\theta))*(x/V _{0})^{2}$

7. anonymous

Then, 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.