At what horizontal distance between the target and the rocket should you launch?

Question

As a science fair project, you want to launch an 850 g model rocket straight up and hit a horizontally moving target as it passes 31 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?


I was trying to use some of the kinematic equations to figure this one out, but I'm always missing some part to the problem or at least that's what I'm thinking. Got any ideas?

Answer 1

Whenever it seems like you are missing some part to a two-part problem, chances are you will need the number that connects the two parts.

The two parts to the problem are the incoming target and the rocket being fired straight up. The common thing connecting these parts is time, which must be the same in both sets of equations. Otherwise, the rocket would miss the target!

For the moving target, use the kinematic equation v * t = x (velocity times time = displacement). x is what you are solving for, so that is an unknown. However, you do know the velocity. But what about time? This is where you need the other part of the problem to figure this out.

In the rocket part of the problem, you can use the kinematic equation 0.5*a*t^2 + v*t = x (0.5 times acceleration times time squared plus initial velocity times time equals displacement). You know the displacement and the initial velocity (zero). To find the acceleration, you will need to use find the net force on the rocket and F = m*a.

Net force = force up (thrust) - force down (gravity)
Net force = 15.0 N - m*g = 15.0 N - 0.850 kg * 9.8 m/s^2
Net force = 15.0 N - 8.33 N = 6.67 N
Net force = m*a
6.67 N = 0.850 kg * a
a = 7.85 m/s^2

Now you can use your rocket equation to find time.
0.5*a*t^2 + v*t = x
0.5*7.85*t^2 + 0*t = 31 m
3.92*t^2 = 31 m
t^2 = 7.91
t = 2.81 s

And finally you can find the distance the target starts from.
v * t = x
15 m/s * 2.81 s = x
x = 42 m

Answer 2

First figure how long it will take the rocket to climb 31m. This you get from the formula s = 0.5*a*t^2, where a is the acceleration. The acceleration will be a = F/m. F is the total force acting on t he rocket, which is thrust - m*g; therefore F = 15.0N - 0.850*g and

a = (15.0 - 0.850*g)/0.850 = 15/0.850 - g = 7.84 m/sec^2 (Newtons are kg*m/sec^2, so you must use kg as the mass.)

The time is then √[2*31 / 7.84] = 2.81 sec. The target will move a distançe of that time times 15 m/s, or

15*2.81 = 42.18 m

This is the distance at which the rocket should be fired.

Answer 3

first of find acceleration of the rocket.

Fnet = ma

there are two forces that at on the rocket, the weight of the rocket and the thrust. Since the rocket moves upward, thrust must be greater than the weigt

thrust - weigh = ma
15N - (.85kg)(9.8m/s^2) = (.85kg)a
6.67N = (.85kg)a
a = 7.847 m/s^2

now you need to find the time it takes the rocket to reach 31m. But first, you need to find the final velocity of the rocket at 31m.

Vf^2 = 2ad + Vi^2
Vf^2 = 2(7.847)(31) + 0^2
Vf^2 = 486.514
Vf = 22.057m/s

now you can find the time
Vf = at + Vi
22.057 = (7.847)t + 0
t= 2.811s

the rocket hit the target at 2.811s. Use the time to find the distance from the lanching point.

x = vt
x = (15)(2.811)
x = 42.165m

I'm not very sure because of the information of the thust. I don't know if its the Fnet, or just the thrust. If it is the Fnet, then to find accleration, take 15N and divide by .85kg. The rest are the same process

Answer 4

i think of the unique velocity is 0. the stairs required are all basic. i'm going that might assist you to work out the sequence of basic steps. you ought to calculate the internet tension appearing on the rocket and the acceleration that internet tension supplies the rocket. Fnet = Fup - Fdown Fnet = m*a Then calculate how long it is going to take the rocket to climb to an altitude of 33m. d = Vo*t + (a million/2)*a*t^2 the place Vo=0 Then locate how some distance the objective's velocity of 15 m/s strikes it for the time of time t.