Let [tex]\vec r(t),\vec v(t),\vec a(t)[/tex] denote the rocket's position, velocity, and acceleration vectors at time [tex]t[/tex].
We're given its initial position
[tex]\vec r(0)=\langle0,0,10\rangle\,\mathrm m[/tex]
and velocity
[tex]\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}[/tex]
Immediately after launch, the rocket is subject to gravity, so its acceleration is
[tex]\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}[/tex]
where [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].
a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,
[tex]\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}[/tex]
[tex]\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}[/tex]
(the integral of 0 is a constant, but it ultimately doesn't matter in this case)
[tex]\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}[/tex]
and
[tex]\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m[/tex]
[tex]\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m[/tex]
[tex]\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}[/tex]
b. The rocket stays in the air for as long as it takes until [tex]z=0[/tex], where [tex]z[/tex] is the [tex]z[/tex]-component of the position vector.
[tex]10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s[/tex]
The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):
[tex]\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}[/tex]
c. The rocket reaches its maximum height when its vertical velocity (the [tex]z[/tex]-component) is 0, at which point we have
[tex]-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)[/tex]
[tex]\implies\boxed{z_{\rm max}=125,010\,\rm m}[/tex]
