A projectile is fired with an initial velocity of \(v\) at an angle of \(\theta\) from the ground. Then its trajectory can modeled by the parametric equations
\(\begin{align*} x &= vt \cos \theta, \\ y &= vt \sin \theta - \frac{1}{2} gt^2, \end{align*}\)

where  \(t\) denotes time and \(g\) denotes acceleration due to gravity, forming a parabolic arch.

Then the maximal area under the trajectory is given by \(k \cdot \frac{v^4}{g^2}\) for some constant \(k\). Find \(k\).



Don't know what to do. got on a track with the integral of y dx with some limits. but ahh idk, solutions and (generous) hints appreciated

 Jul 20, 2020

Consider theta to be given and calculate the time of the flight, ( solve y = 0), and from that, the horizontal distance covered.

This will be the upper limit for the integral that follows.

The parameter is t, so eliminate that between the equations x = and y = to arrive at an equation y = ' a function of x and theta '.

The area under the flight path will be the integral of y wrt x between x = 0 and x = the limit calculated earlier.

The result of that is a function involving v, g and theta. You now have to calculate the value of theta for maximum area, 

(differentiate wrt theta and put that equal to zero etc.).

Post again if you need further help.

 Jul 22, 2020

Here's a ittle more help:

Now differentiate k with respect to \(\theta\) , set to zero and solve for \(\theta\)  Select the solution that gives a value for k leading to the maximal area.

 Jul 22, 2020

41 Online Users