Projectile for PROs

The area under any arbitrary projectile is given by $\frac { 2 }{ 3 } (Max.\quad Height)(Max.\quad horizontal\quad distance)$ . Here, the maximum height is given by $\frac { { u }^{ 2 }sin^{ 2 }{ \theta } }{ 2g }$ and the maximum horizontal distance i.e. the range, is given by $\frac { { u }^{ 2 }sin{ 2\theta } }{ g }$ .

Putting in the given values, the area comes out to be:

$\quad \quad \quad \frac { 2 }{ 3 } (\frac { { u }^{ 2 }sin^{ 2 }{ \theta } }{ 2g } )(\frac { { u }^{ 2 }sin{ 2\theta } }{ g } )\\ \\ \Rightarrow \frac { { u }^{ 4 }sin^{ 2 }{ \theta }sin{ 2\theta } }{ 3{ g }^{ 2 } } \\ \\ \Rightarrow \frac { { u }^{ 4 }sin^{ 3 }{ \theta \cos { \theta } } }{ 3{ g }^{ 2 } }$

Now, for area to be maximum, the derivative of area, must be equal to $0$ . So,

$\quad \quad \frac { dA }{ d\theta } \quad =\quad 0\\ \\ \Rightarrow \frac { d }{ d\theta } \frac { { u }^{ 4 }sin^{ 3 }{ \theta \cos { \theta } } }{ 3{ g }^{ 2 } } \quad =\quad 0\\ \\ \Rightarrow \frac { { u }^{ 4 } }{ 3{ g }^{ 2 } } \frac { d }{ d\theta } sin^{ 3 }{ \theta }\cos { \theta } =\quad 0\\ \\ \Rightarrow \cos { \theta } \frac { d }{ d\theta } sin^{ 3 }{ \theta }\quad +\quad sin^{ 3 }{ \theta }\frac { d }{ d\theta } \cos { \theta } \quad =\quad 0\\ \\ \Rightarrow \cos { \theta } (3sin^{ 2 }{ \theta })(\frac { d }{ d\theta } sin{ \theta })+sin^{ 3 }{ \theta }(-sin{ \theta })\quad =\quad 0\\ \\ \Rightarrow 3sin^{ 2 }{ \theta }cos^{ 2 }{ \theta }\quad -\quad sin^{ 4 }{ \theta }\quad =\quad 0\\ \Rightarrow sin^{ 2 }{ \theta }\quad (3cos^{ 2 }{ \theta }-sin^{ 2 }{ \theta })\quad =\quad 0$

Now, either $sin{ \theta }\quad =\quad 0$ , which won't really form a projectile. Or, $3cos^{ 2 }{ \theta }-sin^{ 2 }{ \theta }\quad =\quad 0\\ \tan ^{ 2 }{ \theta } \quad =\quad 3\\ \tan { \theta } \quad =\quad \pm \sqrt { 3 }$

But since, $\theta <90°$

So, $\tan { \theta } >0\\ \tan { \theta } \quad =\quad \sqrt { 3 } \\ \theta \quad =\quad 60°$

CHEERS!:)

I solved this question a few days back today i got some time so wrote the solution.

Integrate the area within the trajectory.

that will be [ integral f(x) - g(x) dx ] where f(x) is the equation for the trajectory and g(x) is 0 (since on x axis y = 0).I leave the integration to you. Limits are 0 and formula for range.Once done we have got the area as a function of theta (the angle).

now differentiate that w.r.t to theta and set it to zero.

and in the end we get tan theta = sqrt 3. giving theta = pi /3 radians = 60 degrees.

of course we could check whether theta is global maxima or minima by applying second derivative test. (i did n't do it but it suddenly came to my mind.)

then differentiating it again we get an expression in theta but setting theta = 60 we can check whether it is positive or negative if negative theta is global maxima.

really interesting problem integration followed by differentiation.

why not ask the area instead of just angle ?