For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quote a number.

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

In that particular situation, you can prove it like this:

initial velocity is Vo

launch angle is α

initial vertical velocity is

Vv = Vo*sin (α)

horizontal velocity is

Vh = Vo*cos (α)

total time in the air is the the time it needs to fall back to a height of 0 m, so

d = v*t + a*t²/2

where

d = distance = 0 m

v = initial vertical velocity = Vv = Vo*sin (α)

t = time = ?

a = acceleration by gravity = g ( = - 9.8 m/s²)

so

0 = Vo*sin (α) * t + g*t²/2

0 = (Vo*sin (α) + g*t/2) * t

t = 0 (obviously, the projectile is at height 0 m at time = 0s)

or

Vo*sin (α) + g*t/2 = 0

t = - 2*Vo*sin (α) / g

Now look at the horizontal range.

r = v * t

where

r = horizontal range = ?

v = horizontal velocity = Vh = Vo*cos (α)

t = time = - 2*Vo*sin (α) / g

so

r = (Vo*cos (α)) * (-2*Vo*sin (α) / g)

r = - (Vo) ²*sin (2α) / g

To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0.

dr/dα = d[ - (Vo) ²*sin (2α) / g] / dα

dr/dα = - (Vo) ²/g * d[sin (2α) ] / dα

dr/dα = - (Vo) ²/g * cos (2α) * d (2α) / dα

dr/dα = - 2 * (Vo) ² * cos (2α) / g

Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when

cos (2α) = 0

2α = 90°

α = 45°