why does a projectile of 45 degrees have the greatest range?

Answer 1

recall that the formula for range is:
R = (vo)^2 * sin (2*theta) / (2g)
where
vo = initial velocity
theta = angle of release
g = acceleration due to gravity (9.8 m/s^2)

now, if we're given the initial velocity, the value of range will depend on theta,, observing this, if we want to have a max Range, we have to find the max value of sin (2*theta),, but the max value of sine is always 1, therefore, equating this to 1:
sin (2*theta) = 1
2*theta = arcsin 1
2*theta = 90 degrees
theta = 45 degrees

hope this helps~ :)

Answer 2

oops, sorry the formula for range must be
R = (vo)^2 * sin (2*theta) / g

i just remembered that the 2g is for the max height:
h,max = (vo)^2 * [sin (theta)]^2 / 2g

but explanation and proving is still the same~ :)

Answer 3

the greatest sin value of theta is the sin of 45*