Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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
\[\theta = ArcTan \left( \frac{v_0}{\sqrt{2gh+v_0^2}} \right)\]

Answer 5

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