Question

With arbitrary launch angle, the range will be diluted by cos(theta):

x=((v sub-o)cos(theta))t,

while the vertical distance fallen is y, given by

-y=((v sub-o)sin(theta))t-(1/2)g(t^2)

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

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

Answer 1

post the question in mathematics section

Answer 2

Now, we need to find the angle at which x is maximum. Since, at the instant particle lands, vertical distance dropped is h, we'll put y=h. So, the equation becomes

\[-h = v _{0}\sin (\theta)t-\frac{ 1 }{ 2 }g t^{2}\]
On eliminating t from the 1st and 3rd equation, we get
\[-h \cos ^{2}\theta = \frac{ x }{ 2 }\sin 2\theta - \frac{ 1 }{ 2 }g \frac{ x^{2} }{ v _{0} ^{2}} \]
Taking the derivative w.r.t theta on both sides, we'll next put
\[\frac{ dx }{ d \theta } = 0\]
which is the maxima condition for x.
I believe you can do that.