How High Will the Rocket Be When the Fuel Canister Hits the Launch Pad?

[cwalden91]

Homework Statement

During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.30 m/s^2. When it is 235 m above the launch pad, is discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity. (air resistance can be ignored). How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration? What total distance did the canister travel between its release and its crash onto the launch pad?

Homework Equations

V_x=V_ox+a_xt
X=X_o+V_oxt+1/2 a_xt²
V_x²=V_ox²+2a_x(x-x_o)
x-x_o=[(v_ox+v_x)/2)t

The Attempt at a Solution

I have no idea how to go about solving this problem. I've been staring at it for about 30 min. Please help/

 

[Chewy0087]

okay so you know that it accelerates at a steady 3.3m/s^2

so this equation you have here;

Vx2=Vox2+2aS

so here, final velocity^2 = initial velocity^2 + 2 * acceleration * distance travelled, right?

so how about for a start you can find out the speed the canister (and the rocket) were traveling when it reaches 235m.

you have the initial velocity, acceleration and distance, so work out the final velocity and then think about what that means when it's released, and gravity acts on it! (hint: it has to slow down to 0 m/s before it starts falling!)

 

[tomcue]

I would approach this problem by breaking it down into smaller, more manageable parts and using the equations of motion to solve for the unknown variables. First, I would identify the relevant variables in the problem, such as initial velocity (Vox), acceleration (ax), and displacement (x). Then, I would use the given information to solve for the unknown variables.

In this case, we are given the acceleration (3.30 m/s^2) and the displacement (235 m) at which the rocket discards the canister. We also know that the canister is released with an initial velocity of zero and that the only force acting on it is gravity.

Using the equation Vx=Vox+axt, we can solve for the time (t) it takes for the canister to reach the launch pad after it is released. Plugging in the values, we get:

0 = 0 + (3.30 m/s^2)t
t = 0 seconds

This means that the canister hits the launch pad immediately after it is released. Therefore, the height of the rocket when the canister hits the launch pad is equal to the displacement at which it was released, which is 235 m.

To find the total distance traveled by the canister, we can use the equation X=Xo+Voxt+1/2 axt2. Again, the initial velocity is zero and the acceleration is -9.8 m/s^2 (due to gravity). Plugging in the values, we get:

X = 0 + 0 + 1/2 (-9.8 m/s^2)(0 seconds)^2
X = 0 meters

This means that the canister travels a total distance of 0 meters between its release and its crash onto the launch pad.

In conclusion, the rocket is 235 m above the launch pad when the canister is released and the canister travels a total distance of 0 meters before hitting the launch pad.