How high up will the rocket impact? If I fire a rocket at a building $20,000$ feet away at an initial velocity of $1000 f/s$, and an angle of $20^\circ$, how high up will it impact the building?
I resolved the triangle and found, after much work, that the rocket impacts in 21.3 seconds at a final velocity of 999.17 at an angle of -20 degrees (but the book uses 160, but i can see this).
But, I don't know how to go about getting the answer to how high up it impacts.  Can you assist?
 A: The rocket never reaches the building.  From resolving the vertical component of the velocity, we can see how far it would get if there were no building in the picture.  That distance is $$1000\cos(20)\times 2 \times \frac{1000 \times \sin(20)}{g}=19,981 ft$$
Assume there is no underground impact, that is.... If you allow for a target that is 20,000 ft away, but where the impact can be below the level of firing, then $h=vt-\frac{gt^2}{2}=-6.93$ft where $t=\frac{20,000}{u}$, where $u=1000\cos(20)$ and $v=1000\sin(20)$.
A: Let's assume that this is a problem in projectile motion--i.e. all the force is imparted to the rocket at the start and afterwards the rocket has constant downward acceleration of $32.2\ \mathrm{ft}/\mathrm{s}^2$. Let's also assume that the 20° is the angle of inclination from the horizontal. (These is not clear from your problem statement.)
Resolve the initial velocity vector into its horizontal and vertical components. Use the horizontal component and the fact that the horizontal velocity remains constant to find the time of impact. This does not require "much work"--just solve the equation
$$20000 = 1000\cos(20°)\cdot t$$
or
$$20000 = 1000\cos(20°)\cdot\Delta t$$
depending on your notation. You are correct that the time of impact is 21.3 seconds, to three significant digits, but it is not at all clear that this is the desired precision. Your given data does not seem to support that many significant digits. Are you sure that is the correct precision?
Then use the vertical component of the initial velocity and the equations of constant acceleration to find the final displacement of the rocket. You are given the time (you just found it), the initial velocity (the vertical component), and the acceleration ($-32.2\ \mathrm{ft}/\mathrm{s}^2$), so find the displacement, using the equation
$$s = ut + \frac 12at^2$$
or
$$\Delta x = v_i\Delta t + \frac 12a(\Delta t)^2$$
depending on your notation. That gives the result $-19.4$ feet, again with questionable precision. So the rocket ends up lower than it started, which is possible if it starts from an elevated platform or hill or some such.
But again, the precision in all this is questionable, so this answer is debatable but is near the starting level.
