Question

You launch a model rocket from ground level. it moves directly upward with a constant acceleration of 80.5 m/s2 for 1.40 seconds, at which point it runs out of fuel. assuming air resistance on the rocket is negligible, what is the maximum altitude (above the ground) achieved by the rocket

Answer 1

The model rocket reaches a maximum altitude of 645.09 meters above ground level after it runs out of fuel, calculated using kinematic equations for the coasting phase of the flight.

Step-by-step explanation:

Calculating the Maximum Altitude of the Model Rocket

To calculate the maximum height achieved by a rocket, we can use kinematic equations. The rocket's motion occurs in two stages: ascent with fuel and the coasting motion upwards until gravity stops it. First, we will find the rocket's final velocity when it runs out of fuel, and then we will use this value to find the height reached during the coasting phase.

Stage 1: Ascent with Fuel

The final velocity (v) at the end of the burn can be found using the equation:
v = u + at, where u is the initial velocity (0 m/s, as it starts from rest), a is the acceleration (80.5 m/s2), and t is the time with fuel (1.40 s).

Calculating final velocity: v = 0 + (80.5 m/s2 * 1.40 s) = 112.7 m/s.

Stage 2: Coasting to Maximum Height

Once the fuel is exhausted, the rocket will continue to ascend under its own inertia until gravity brings it to a stop. The maximum height during this coasting phase can be found using the equation:
v2 = u2 + 2as (where s is displacement, and a is now the acceleration due to gravity, which is -9.80 m/s2 because it acts downwards).

At maximum height, the final velocity will be 0 m/s. Plugging in the values, we get:
0 = (112.7 m/s)2 + 2(-9.80 m/s2)s, which simplifies to s = (112.7 m/s)2 / (2 * 9.80 m/s2) = 645.09 m.

Therefore, the maximum altitude above the ground achieved by the rocket is 645.09 meters.

Note: The sum of the heights during the acceleration phase and the coasting phase is the total maximum height. However, since the initial velocity for the coasting phase is the final velocity of the acceleration phase, only the coasting phase needs to be considered for the final height calculation.

Answer 2

If the rocket starts at rest, then

`a=\frac vt\iff80.5\,(\mathrm m)/(\mathrm s^2)=\frac v{1.40\,\mathrm s}\implies v=113\,(\mathrm m)/(\mathrm s)`

Then in the 1.40 seconds during its initial ascent, the displacement over this period is given by

`x=x_0+\frac12(v+v_0)t\iff x=\frac{\left(113\,(\mathrm m)/(\mathrm s)\right)(1.40\,\mathrm s)}2=78.9\,\mathrm m`

Once fuel runs out, the only force acting on the rocket is that due to gravity, so the rocket's acceleration drops to a constant
`-9.81\,(\mathrm m)/(\mathrm s^2)`. The rocket's displacement is then captured by

`x=x_0+v_0t+\frac12at^2\iff x=78.9\,\mathrm m+\left(113\,(\mathrm m)/(\mathrm s)\right)t+\frac12\left(-9.81\,(\mathrm m)/(\mathrm s^2)\right)t^2`

`\implies x=78.9\,\mathrm m+\left(113\,(\mathrm m)/(\mathrm s)\right)t+\left(-4.91\,(\mathrm m)/(\mathrm s^2)\right)t^2`

You can find the maximum displacement from here by completing the square:

`x=729\,\mathrm m+\left(-4.91\,(\mathrm m)/(\mathrm s^2)\right)(t-11.5\,\mathrm s)^2`

This tells you that a maximum displacement/altitude of about
`729\,\mathrm m` is achieved by the rocket at
`t\approx11.5\,\mathrm s`.