A projectile near the earth's surface experiences a constant gravitational acceleration of
r''(t)=a(t) = (0,0, -g)
suppose r(0) = (0,0,0) and r'(0) = (a,b,c), where c>0. solve for the trajectory of the projectile r(t), and find where its speed is minimized.
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r(t) = (x(t), y(t), z(t))
r'(t) = (x'(t), y'(t), z'(t))
r''(t) = (x''(t), y''(t), z''(t)) = (0, 0, -g)
x''(t) = 0
x'(t) = a
(since we integrate x''(t) to get x'(t), x'(t) must be a constant and since x'(0) = a, that constant must be a)
x(t) = at + C where C is a constant
Since x(0) = a(0) + C = C, and we know that x(0) = 0, we have C = 0
Therefore, x(t) = at
y''(t) = 0
y'(t) = b
(since we integrate y''(t) to get y'(t), y'(t) must be a constant and since y'(0) = b, that constant must be b)
y(t) = bt + C' where C' is a constant
Since y(0) = b(0) + C' = C', and we know that y(0) = 0, we have C' = 0
Therefore, y(t) = bt
z''(t) = -g
z'(t) = -gt + C'' where C'' is a constant
z'(0) = -g(0) + C'' = C'', but we are given that z'(0) = c, therefore C'' = c
z'(t) = -gt + c
z(t) = -(1/2)gt^2 + ct + C''', where C''' is a constant
z(0) = -(1/2)g(0)^2 + c(0) + C''' = C''', but we are given that z(0) = 0, therefore C''' = 0
z(t) = -(1/2)gt^2 + ct
r(t) = (x(t), y(t), z(t)) = (at, bt, ct - (1/2)gt^2 )
Speed at time t = s(t) = |r'(t)| = |(x'(t), y'(t), z'(t))| = sqrt( a^2 + b^2 + (c - gt)^2) )
= sqrt( (a^2 + b^2 + c^2) - 2cgt + g^2(t^2) )
In order to minimize speed, s(t), we set its first derivative to zero and solve to find a critical point.
s'(t) = (1/2) ( (a^2 + b^2 + c^2) - 2cgt + g^2(t^2) )^(-1/2) ( - 2cg + 2(g^2)t)
s'(t) = ( (a^2 + b^2 + c^2) - 2cgt + g^2(t^2) )^(-1/2) ( - cg + (g^2)t)
Solving for s'(t) = 0, we have
0 = ( (a^2 + b^2 + c^2) - 2cgt + g^2(t^2) )^(-1/2) ( - cg + (g^2)t)
0 = (-cg + (g^2)t) (assuming a, b are not both 0)
cg = (g^2)t
t = c/g
(If a and b are both 0, then s'(t) is not defined at t = c/g, but that still makes t = c/g a critical point to evaluate for minimization - note that s(c/g) = 0 in this case so this must be a minimum).
Thus at time t = c/g, the speed is minimized. The minimum value can be calculated as
s(c/g) = |r'(c/g)|
= sqrt( a^2 + b^2 + (c - g(c/g))^2) )
= sqrt( a^2 + b^2 + (0)^2) )
= sqrt( a^2 + b^2 )
The position of the particle when it attains minimum speed can also be calculated as
r(c/g) = (a(c/g), b(c/g), c(c/g) - (1/2)g(c/g)^2 )
= (ac/g, bc/g, c^2/g - (1/2)g(c^2/g^2) )
= (ac/g, bc/g, c^2/g - (1/2)(c^2/g) )
= (ac/g, bc/g, c^2/(2g))