Area under projectile's curve

Question

+1

343

4

+72

Here is a problem from lecture notes:

Projectile is fired with initial velocity v from the ground, at an angle of theta. Its trajectory is parameterized:

x = vt cos theta

y = vt sin theta - 1/2 gt^2

t = time, g = acceleration due to gravity.

A parabolic arch is formed by the projectile. The maximum area under the trajectory is given by k x v^4 / g^2 where k is a constant. Find k.

jsaddern Dec 24, 2021

Melody · Answer 1 · 2021-12-24T01:29:40

This question doesn't quite make sense to me.

if v is a set constant but theta is variable (which is not stated)

then I got the max area when theta = pi/6

The max area would be \(\frac{\sqrt3}{24}*\frac{v^4}{g^2}\)

I have not checked my working and it could easily be wrong but it is at least in the form that is asked for.

If you want to ask questions feel free to do so.

Melody · Answer 2 · 2021-12-25T02:58:29

This answer seems less likely and more prone to error but on a second attempt I got

\(Max = \frac{(33-9\sqrt{13})\sqrt{6\sqrt{13}-18}}{6}\:\displaystyle \frac{v^4}{g^2}\)

Do you know the actual answer by any chance?