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:
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.
But we already know `dx/(dt)` and x at impact, so we simply substitute:
This gives us a negative velocity, as we expected before:
So now we need to find the magnitude of the velocity. This takes into account both the horizontal and vertical components.
Substituting, we have:
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:
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.