Why is the greatest range of a projectile produced when the angle is 45 degrees?

Answer 1

Suppose the projective is launched, over a level surface, with initial velocity u, which makes an angle x with the horizontal. Then 0 <= x <= 90 degrees.Ignore the effects of all forces other than gravity,

Ananlyse the motion in the vertical direction.

The vertical component of u is u*sin(x), and the projectile is in the air until its vertical displacement returns to 0.

That is t is given by

0 = u*sin(x)*t - 1/2*g*t^2 where g is the acceleration due to gravity.

ie 0 = t*{u*sin(x) - 1*2*g*t}

that is t = 0 or u*sin(x) - 1/2*g*t

ignoring t = 0 since that is the initial position,

t = 2*u*sin(x)/g.

Now consider the horizontal motion.

The horizontal component of the initial velocity is u*cos(x) and [since we have chosen to ignore aerodynamic drag], this is a constant.

Therefore, horizontal distance covered in t units of time = u*cos(x)*t units of length.

Substituting for t gives

range = u*cos(x)*2*u*sin(x)/g = (u^2/g)*2*cos(x)*sin(x)

= c*2*cos(x)*sin(x) where c is a constant

= c*sin(2x) [using the double angle formula].

Now d(range)/dx = 2c*cos(2x) must be 0 at the maximum.

[It cannot be the minimum since x = 90 degrees gives a minimum range of 0].

therefore cos(2x) = 0

therefore 2x = 90 degrees and so x = 45 degrees.