Question

Can you combine these equations and seek the angle theta that theoretically maximizes the range x, when launched at height h above the landing field?

Equations:
x=(v cos(theta)) t
-y=(v sin(theta)) t - (1/2) g (t^2)

Hint: this procedure is relatively easy on a horizontal surface, where y = 0; there the maximum range occurs when theta = 45 deg.

Answer 1

firstly we'll have to put\[y = h\]
then substitute t from the 1st equation in tthe second

Answer 2

So do I isolate t in the first equation? And get

\[t=\frac{ x }{ v \cos (\theta) }\]

Answer 3

yes

Answer 4

which will give me
\[y=h=(v \sin (\theta))\frac{ x }{ v \cos(\theta) }-(\frac{ 1 }{ 2 }) g (\frac{ x }{ v \cos(\theta) })^{2}\]

Answer 5

therefore we get \[h=\frac{ x \sin(\theta) }{ \cos(\theta) }-\frac{ g x }{ 2 v \cos(\theta) }\]

Answer 6

then how do I find out theta from there?

Answer 7

do you know maxima condition ?

Answer 8

well thats just a hint of the problem. the mad range occurs when theta is 45 deg

Answer 9

On putting dx/dtheta = 0 , and simplification , I came up with an eq. that again expresses x as some fuction of theta.
So, we can again differentiate and put dx/dtheta = 0 to come up with the theta.

Answer 10

is the ans in terms of g and v nought ?

Answer 11

My goodness, this isn't easy. I have a suggestion for you: convert the above problem into projectile motion on inclined plane. Derivation of inclined plane max range , theta, etc must have been taught to you ( or google it ) It will help you solve this question.

(I get bad trigo eqs when I attempt solving them :( will contact you, if I'm able to get a neat solution later)

Answer 12

I am getting bad trig functions too. That's why I'm so confused. but I don't think i learned the inclined plane max range.