Verified answer

r' = integral r'' dt = integral (0,0,-g) dt = (A,B, -gt + C)

A, B, C = constants

r'(0) = (a,b,c) = (A,B,-g(0) + C)

a = A

b = B

c = C

so r'(t) = (a,b,-gt + c)

r = integral{ r' dt} = integral (a,b,-gt + c) dt

= (at + A', bt + B', -gt^2 / 2 + ct + C')

where A', B', C' are constants

r(0) = (0,0,0) = (A', B', C')

A' = 0

B' = 0

C' = 0

so

r(t) = (at, bt, -gt^2 / 2 + ct).........[Ans.]

which is exactly what you would suspect. The projectile starts out with velocity in the x and y directions of (a,b), at a time t later, the projectile has traveled at, and bt in the x and y directions, respectively (i.e. distance = velocity * time). In the x and y directions there is no acceleration (no change in speed), so we just have them constantly moving at their given rates of a and b.

In the z direction we have r = -gt^2 / 2 + ct, which is the usual constant acceleration equation for a projectile that starts at an initial position of zero, which it does.

To find when its speed is minimized, notice r' = velocity

| r' | = speed = sqrt{a^2 + b^2 + (c - gt)^2}

minimize speed by seeking d|r'| / dt = 0

d|r'| / dt = (1/2){a^2 + b^2 + (c - gt)^2}^{-1/2} { 2 (c - gt)(-g) } = 0

This amounts to the numerator vanishing

c - gt = 0

---> t = c / g .....................[Ans.]

is when the speed is minimized

Minimizing the speed |r'(t)| also minimizes the square of the speed |r'(t)|²

r'(t) = r'(0) + ∫[0,t] r"(t) dt = (a, b, c) + (0, 0, -gt)

r'(t) = (a, b, c-gt)

|r'(t)|² = a² + b² + (c - gt)²

This is clearly minimized when c=gt or t = c/g

For the trajectory, integrate:

r(t) = r(0) + ∫[0,t] r'(t) dt = (0, 0, 0) + (at, bt, ct - gt²/2)

r(t) = (a,b,c)t - (0, 0, g/2)t²