Let's first see a graph of the motion, to better understand what is going on.

We can see that the rocket hits the ground again somewhere around *x* = 9.5 km. At this point, the horizontal velocity is positive (the rocket is going left to right) and the vertical velocity is negative (the rocket is going down).

"*V*(*x*) = *x*" means that as *x* increases, the horizontal velocity also increases with the same number (different units, of course). So for example, at *x* = 2 km, the horizontal speed is 2 km/min, and at *x* = 7 km, the horizontal speed is 7 km/min, and so on.

To calculate the **magnitude** of the velocity when the rocket hits the ground, we need to know the vertical and horizontal components of the velocity at that point.

(1) **Horizontal velocity.** We just need to solve the following equation to find the exact point the rocket hits the ground:

`x-x^3/90=0`

Factoring gives:

`x-x^3/90=x(1-x^2/90)`

And solving for 0 gives us `x = 0`, `x = -3sqrt(10)`, `x = 3sqrt(10)`

We only need the last value, `x = 3sqrt(10)~~ 9.4868\ "km"` (This value is consistent with the graph above).

So the horizontal speed when the rocket hits the ground is 9.4868 km/min (since *V*(*x*) = *x*).

(2) **Vertical velocity. **We now need to use **implicit differentiation** with respect to *t* (not *x*!) to find the vertical velocity.

`y=x-x^3/90`

`(dy)/(dt)=(dx)/(dt)-1/30x^2(dx)/(dt)`

But we already know `dx/(dt)` and* x *at impact, so we simply substitute:

`(dy)/(dt)` `=(9.48683298)-` `1/30(9.48683298)^2(9.48683298)`

This gives us a negative velocity, as we expected before:

`(dy)/(dt)=-18.97366596`

So now we need to find the magnitude of the velocity. This takes into account both the horizontal and vertical components.

`"Magnitude"=sqrt(((dx)/(dt))^2+((dy)/(dt))^2`

Substituting, we have:

`sqrt((9.48683298)^2+(-18.97366596)^2)` `=21.21320344`

Velocity has **magnitude** and **direction**. Now for the direction part.

`"angle of motion" = arctan\ (dy"/"dt)/(dx"/"dt)`

Substituting our vertical and horizontal components, we have:

`arctan\ (-18.97366596)/9.48683298` `=-1.107148718`

In degrees, this is equivalent to

`-1.107148718 × 57.25578` ` = -63.3907^@`

We can see that this answer is reasonable by zooming in on the portion of the graph where the rocket hits the ground (with equal-axis scaling):

So in summary, the velocity of the rocket when it hits the ground is 21.2 km/min in the direction `63.4^@` below the horizontal.