# 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?

• What is the radius and surface gravity of the planet where this question is set? Also, you say "rocket", what is the thrust, initial mass and fuel consumption rate? – Joffan Oct 1 '17 at 0:16
• Is it implied in the context 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 9.81 m/s^2? Also, how is this a calculus problem? It seems more like mathematical physics and/or Precalculus. – Rory Daulton Oct 1 '17 at 0:26
• I so totally agree....how is this a calculus problem!! It is NOT. However, I'm in a calculus class with an ego-maniac for an instructor who wants to make this into a hybrid. I signed up for a Calculus class.....got a hybrid. – user163862 Oct 1 '17 at 0:45

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 $v=1000\sin(20)$.

• Our calculations don't match. I get that if there were no building the rocket would go 21.375 (1000 cos(20) = 20085.93 ft. – user163862 Oct 1 '17 at 0:50
• My calcs show that the rocket DOES reach the building. – user163862 Oct 1 '17 at 1:27
• @user163862 Hmmm...wanna check again? Basically, Rory's answer and mine are the same in method, except I use an extra digit of precision for $g$ (32.17 instead of 32.2). For $t$, if you take an extra level of precision, it is 21.28. The range of the rocket is so close to 20,000 that precision really matters to determine above or below ground. I think all three of us are "correct", upto precision levels. Did you use 32 ft/s^2 for g? – Mathemagical Oct 1 '17 at 2:09
• Well, the book says it impacts the building (that's what the question is all about) and 2 other physics people say it impacts the building and my calculations say the total time of the rocket should be 21.375 s if there is no building. So, at 21.3 s, it will hit the building. – user163862 Oct 1 '17 at 2:41
• I used 32 ft/s/s – user163862 Oct 1 '17 at 2:43

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.

• Regarding correct precision....it's a calculus class and the instructor states nothing about precision. His answers to the quesitons of how long will it take to hit the building is 21.3 seconds. What will it's impact velocity be: 999.17 ft/s. it's unclear to me how many digits of precision he's looking for. – user163862 Oct 1 '17 at 1:23
• His answer (without any explanation) says 25.99 feet for the answer. – user163862 Oct 1 '17 at 1:24
• He doesn't state anything about elevation of where the rocket is fired from, so I think it's safe to assume ground level. – user163862 Oct 1 '17 at 1:26