A500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. the rocket engine, when it is fired, exerts an 8.0 n thrust on the rocket. your goal is to have the rocket pass through a small horizontal hoop that is 20 m above the launch point. at what horizontal distance left of the hoop should you launch?

x = 7.62 m

Explanation:

First we need to calculate the weight of the rocket:

W  = mg

we will use the gravity as 9.8 m/s². We have the mass (500 g or 0.5 kg) so the weight is:  

W  = 0.5 * 9.8 = 4.9 N

We know that the rocket exerts a force of 8 N. And from that force, we also know that the Weight is exerting a force of 4.9. From here, we can calculate the acceleration of the rocket:

F - W = m*a

a = F - W/m

Solving for a:

a = (8 - 4.9) / 0.5

a = 6.2 m/s²

As the rocket is accelerating in an upward direction, we can calculate the distance it reached, assuming that the innitial speed of the rocket is 0. so, using the following expression we will calculate the time which the rocket took to blast off:            

y = vo*t + 1/2 at²

y = 1/2at²

Solving for t:

t = √2y/a

t = √2 * 20 / 6.2

t = √6.45 = 2.54 s

Now that we have the time, we can calculate the horizontal distance:

x = V*t

Solving for x:

x = 3 * 2.54 = 7.62 m            

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2