bit rusty on this can some1 quickly explain (1 Viewer)

[lmao1010]

ive forgotten the logic behind y launching a proj from 30 degree and 60 degree will result in the same range. i know it is because both angles are equidistant from 45 deg, but I feel I'm missing smthg so can someone expand / clarify.

 

[username_2]

ive forgotten the logic behind y launching a proj from 30 degree and 60 degree will result in the same range. i know it is because both angles are equidistant from 45 deg, but I feel I'm missing smthg so can someone expand / clarify.

Well this is true. One way you can think about it is that at 45 degrees, a projectile has a maximum range. So as I deviate further from 45 degrees on either end, the range changes are going to be equivalent. (But really this has to do with a more mathematical reason where the angle is limited between 0 and 90 degrees results in the max range being at 45 and anything on either end is symmetrically lower).

 

[yanujw]

Suppose a projectile is launched with initial velocity v and angle A
Initial y velocity: Vsin(A)
Time taken to reach peak height (using v=u+at) :
Total time of flight:
Total horizontal range (by the fact that horizontal velocity is constant):

The value of A that maximises this is the value that maximises sinAcosA. Which is 45 degrees.
But also, sinAcosA is symmetrical about 45 degrees. This can be proven by showing the equivalence of sin(45+B)cos(45+B) = sin(45-B)cos(45-B).