This does intuitively make sense. #45^@# is the angle half-way from 0 and a right angle, so this should maximise both vertical distance and horizontal height.
Nonetheless, we can also prove this using a bit of algebra and calculus. Even if you can't do the second bit, first bit should make sense. The task is:
Consider this diagram.
Prove that the distance #r=-(2u^2sinthetacostheta)/g#
Still here? Lets go.
Let's say we are firing a projectile with a velocity of #u# at an angle #theta#, and we want to find the horizontal range #r#. The particle is accelerating to the ground with accelerating #g#.
Considering vectors, the vertical velocity of the particle is #usintheta#.
Find the time it takes for the particle to return to the ground if we throw it up and that velocity.
#s=0# It's returning to the start.
#u=usintheta#
#v="/"# we don't care what v is
#a=-g# since we've chosen up to be the positive direction, this means down must be negative.
#t=t# this is what we want to find.
Using the equation #s=ut+1/2at^2#
#0=utsintheta+1/2g t^2#
#0=1/2t(g t+2usintheta)#
#t=(-2usintheta)/g#
Now, consider the horizontal motion
#s=r# this is what we want to find
#u=ucostheta#
#v="/"#
#a=0#
#t=-(2usintheta)/g#
since #a=0, s=ut#
#r=-ucostheta*(2usintheta)/g#
#r=-(2u^2sinthetacostheta)/g#
Now, as an extra challenge, can you prove that the angle that gives the maximum r is #pi/4 (45^@)#?
We can use some trig identities to help us with this expression, then differentiate to find the maximum angle.
Since #2sinxcosx=sin2x#;
#r=-u^2/gsin(2theta)#
Now, differentiate both sides wrt #theta#. This gives us the gradient function. When this is equal to zero, we will find the maximum value.
Since #d/dx(sinkx)=kcoskx#;
#(dr)/(d theta)=-(2u^2)/gcos(2theta)#
Let #(dr)/(d theta)=0#
#-(2u^2)/gcos(2theta)=0#
#cos(2theta)=0#
We want an angle between 0 and a right angle, so we want a value in the range #0<=theta<=pi/2#
#2theta=pi/2" or " 90^@#
#theta=pi/4" or " 45^@#
