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Help w/Max height

 
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Sep12-07, 06:57 PM   #1
 
Question

Help w/Max height


1. The problem statement, all variables and given/known data

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 49.0 m/s . The acceleration period lasts for time 8.00s until the fuel is exhausted. After that, the rocket is in free fall.

2. Relevant equations

How do i start? What should i do?

3. The attempt at a solution

Dont know where to start...
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Sep12-07, 07:05 PM   #2
 
On these type of problems, you must always interpret the problem and find out the given information first.

The problem states that the rocket is "Initially rest at ground", this means that

[tex]v(distance = 0) = 0[/tex]

meaning initial velocity is equal to 0.

...

I just realized, what is the question asking for? time of free fall? distance? lol

[EDIT]Oh sorry, I just looked at the title, the max height

We know the constant acceleration 49.0m/s^2, which this particular acceleration lasts for 8 seconds AND we also know the gravity (-9.81m/s^2) constantly acts on the rocket as it propels upwards.

And finally, we know the velocity is equal to 0 when the object reaches its highest point.

For such type of questions, you need two equations to model this problem.

First, use the equation

[tex]d = d_0 + v_0 * t + \frac{1}{2}a_r * t^2[/tex]

[tex]a_r = acceleration of rocket - gravitational acceleration[/tex]

We know the initial velocity and distance is equal to 0, so forget about those quantities.

To calculate the total distance of rocket up until 8 seconds, then find the velocity of the rocket, at the point which it stops accelerating by

[tex]v = v_0 + a_r * t[/tex]

Now you need another equation to model the velocity of the rocket after it stops accelerating. The only acceleration acting upon this rocket is gravity now. Use the equation given before again:

[tex]d = d_0 + v_0 * t - \frac{1}{2}a_g * t^2[/tex]

[tex]a_g = 9.81m/s^2[/tex]
Your initial distance would be the distance travelled by the rocket until it stops accelerating, and you've just calculated the velocity at that instant too.

Now you need to use the information which the velocity is equal to 0 when it is at maximum height.

[tex]v = v_0 - gt[/tex]

Solve for t, plug it into the equation above, then voila. You got the answer
Sep12-07, 08:02 PM   #3
 
Thanks!
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