As a science fair project, you want to launch an 600 g model rocket straight up and hit a horizontally moving?

Question

As a science fair project, you want to launch an 600 g model rocket straight up and hit a horizontally moving target as it passes 35 m above the launch point. The rocket engine provides a constant thrust of 15.0 N. The target is approaching at a speed of 15 m/s. At what horizontal distance between the target and the rocket should you launch?

Answer 1

This is a good one. As you know f = Ma most of the time. And I'm guessing most answers will use this. But that's not the case for rockets.

The force-accelerated mass M = m + m(t); where m is the fixed body mass and m(t) is the fuel mass. m(t) is a function of time as the fuel burns away at a rate of dm(t)/dt < 0, which means losing fuel mass over time of burn.

For your rocket, f = dP/dt = d(Mv)/dt = dM/dt v + M dv/dt is in effect; where dM/dt = d(m + m(t))/dt = dm(t)/dt; so that f = dm(t)/dt v + (m + m(t)) dv/dt = dm(t)/dt v + m(t) dv/dt + m dv/dt = dm(t)/dt v + (m(t) + m) dv/dt = 15 Newton (preferable units are kg-m/sec^2). P = Mv is the linear momentum of your rocket and dP/dt = f is the change in that momentum over time, which we define as force.

Thus, a <> f/M, as you are probably suggesting. In fact, for the kind of intercept precision you'd need, a = dv/dt = [f - dm(t)/dt v]/[m(t) + m]. Notice that this latter becomes a = f/M if m(t) = constant so that dm(t)/dt = 0. But, alas, that's not the way rockets work, they lose most of their initial mass by buring away that fuel. That has to be taken into account...especially when trying to intercept another moving body.

If you are really going to do this at a science fair, you need to do more physics. You will also need to take air drag into account because that increases as the square of the velocity. And your intercept velocity will be significant. Finally, guidance, what will keep your rocket on the straight and narrow? What if there are wind gusts?

If you are just doing an incomplete think piece, find the time to climb t for getting to 35 m above the launch site. That can be found from h = 35 m = 1/2 at^2; where a = f/M = 15/.6. Solve for t = sqrt(2ah). Then the rocket needs to be launched when the target is D = vt = 15*t meters from the impact point above the launch pad.

But, and this is a big BUT, you'll likely miss the target if you fail to include the burn rate and other factors I've mentioned earlier.

Answer 2

Model Rocket Science Fair Projects

Answer 3

It would be a bit more accurate if you knew how much fuel the motor burned (how many g/s) because it does change things slightly (the mass of the rocket and, therefore, it's acceleration change). But maybe the numbers here are small enough that it won't make much difference.. Here's the calculation.

A .6 kg rocket has a force of
F=ma=.6*9.8=5.88 N acting on it from gravity. That leaves
15 - 5.88 = 9.12 N of total thrust for acceleration. The acceleration will be (a=F/m)
a = 9.12/.6 = 15.2 m/s² and, since d= (1/2)at² then t = √(2d/a) and
t = √(2*35/15.2) = 2.146 sec will be the time required to get to 35 m. In 2.146 sec. the 'target' will travel
2.146*15 = 32.19 meters and that's your answer.

PS. Lots of luck keeping that rocket moving in a 'straight' line.

Doug