Solving 2 equations (Projectile motion)

Asked Dec 15 '14 at 19:08

Active Dec 15 '14 at 19:53

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The question is from Physics, but all I need is help on solving maths. So basically, i am trying to find out the optimal angle for projectile motion from a certain height and I end up with these two equations:

$$0=h + R \tan \theta - R^2 \frac g{2u^2}(1 + \tan^2 \theta) \tag1$$

$$R= \frac{g}{u^2}cot \theta \tag2$$ where, h = height of the tower

R= maximum distance travelled by the stone (Range)

g=gravitational constant

$\theta$ = angle of the projection

Can anyone help with how do I solve this two equations to obtain this:

$$\theta = \arctan \left(\frac u{\sqrt{u^2+2gh}} \right)$$

Thanks. :)

P.S.: I did tried to solve by substituting R in equation (1), but my final answer way very long and very complicated.

edited Dec 15 '14 at 19:53

asked Dec 15 '14 at 19:08

aavatar

Just replace all occurencies of $R$ in the first equation using the second equation, then solve the quadractic equation with respect to $\tan ^2\theta$. – TZakrevskiy Dec 15 '14 at 19:11
@TZakrevskiy I did it the same way.. but it was way too complicated.. – aavatar Dec 15 '14 at 19:12
Look at http://math.stackexchange.com/questions/571/what-is-the-optimum-angle-of-projection-when-throwing-a-stone-off-a-cliff – Ross Millikan Dec 15 '14 at 19:13
my result is also not yours, are you sure that you have made no typo? – Dr. Sonnhard Graubner Dec 15 '14 at 19:35
@Dr.SonnhardGraubner I pretty much sure... there is not typo – rndflas Dec 15 '14 at 19:43
do you have a link of this problem? – Dr. Sonnhard Graubner Dec 15 '14 at 19:44
@Dr.SonnhardGraubner I am so sorry!! yes there is an typo. Really sorry.. just realized now!! Thank you very very much.. I jst editted the question. – aavatar Dec 15 '14 at 19:52
Since $\cot \theta = 1/\tan \theta$, everything cancels out nicely. You shouldn't be getting anything complicated. – Braindead Dec 15 '14 at 20:07
I suspect there is an error somewhere: $R$ is length, yet $\frac{g}{u^2}$ is $\frac{1}{\text{length}}$, hence the second equation has little sense. – TZakrevskiy Dec 15 '14 at 20:16

Solving 2 equations (Projectile motion)

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