Proving 45 degrees is optimal value for greatest horizontal range

[fengwater]

I'm still doing grade 12 physics, so my physics knowledge is VERY limited.

So I have a question about proving that a projectile launched 45 degrees above horizontal gives the greatest horizontal displacement.

I have to ideas to go about proving this:

1. Subbing Equations
I have currently made 2 equations that I can think of:
Vertical Component: 0=V$$_{}y1$$-4.9t$$^{}2$$
Horizontal Component: t=d$$_{}x$$/V$$_{}x1$$

I subbed equation 2 into equation 1. However, I got stuck at 0=V$$_{}y1$$V$$_{}1x$$-4.9d$$_{}x$$
Where's the third equation?

2. Finding the horizontal displacement of 44 degrees and 46 degrees, and compare their expected lower horizontal displacement to that of 45 degrees. Or basically just find the time the projectile is moving, just solving for t when vertical displacement is 0.


Considering my grade level, is solution 1 possible, or should I use solution 2?

Thanks!

 

[Dschumanji]

Using substitution and algebra is the best way for a proof.

Start by writing down the equations for the position of the projectile. For simplicity let the projectile start at the origin.

The most advanced thing you need to know to do the algebraic proof is the quadratic formula and a knowledge of trig identities.

 

[prateek_34gem]

You should always solve the projectile problem like this :

Let it be projected at angle 'x' with horizontal. Then at instantaneous time 't'.
Let initial velocity be 'a'. :-


Along X- Direction

u=acosx
a=0
v=acosx
s=acosxt

Along Y- Direction

u=asinx
a= -g
v=asinx-gt
s=asinxt-1/2gt²

Now when it reaches the ground again after projectile it's Y-coordinate will be zero.
i.e displacement along y - axis is zero.

so,

asinxt-1/2gt²=0
asinx=1/2gt
t=2asinx/g

Now range is basically your displacement along X-direction.
putting the value of t in acosxt.

Range = a²2cosxsinx/g
= a²sin2x/g

Now for maximum range sin2x should be maximum i.e. =1
sin2x=1
sin2x=sin90
2x=90
x=45
so , for maximum range x = 45