Rocket Acceleration and Free Fall Homework Help

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In summary, the rocket goes from rest to a maximum height in 8 seconds, and then falls back to the ground.
  • #1
Robdog
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Homework Statement



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.

Homework Equations



How do i start? What should i do?

The Attempt at a Solution



Dont know where to start...
 
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  • #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 traveled 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
 
Last edited:
  • #3
Thanks!
 

1. What is the definition of "max height" in physics?

The maximum height, also known as the apex or peak, is the highest point reached by an object in its motion, such as a projectile or a roller coaster.

2. How is max height calculated?

The maximum height is calculated using the equation: H = (V2sin2θ)/2g, where H is the maximum height, V is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

3. Can you provide an example of calculating max height?

Sure! Let's say a ball is thrown with an initial velocity of 20 m/s at an angle of 45 degrees. Using the equation above, we can calculate the maximum height: H = (202sin245)/2*9.8 = 20.4 meters.

4. How does air resistance affect max height?

Air resistance, or drag, can decrease the maximum height of an object by slowing it down and reducing its initial velocity. This is because air resistance acts in the opposite direction of the object's motion, causing it to lose energy and not reach its full potential height.

5. Is max height the same as altitude?

No, max height and altitude are not the same. Altitude refers to the height above sea level, while max height is the highest point reached by an object in its motion. Altitude is a fixed measurement, while max height can vary depending on the initial conditions and forces acting on the object.

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