Physics, 04.07.2020 14:01 ciel8809

A model rocket is fired vertically and ascends with a constant vertical acceleration of 4.00 m/s2 for 5 s. Its fuel is then exhausted and it continues as a free-fall particle.(a) What is the maximum altitude reached?(b) What is the total time elapsed from takeoff until the rocket strikes the Earth?

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Answer from: thuzar

a) The rocket reaches a maximum height of 70.394 meters, b) The total time elapsed from takeoff until impact is 10.829 seconds.

Explanation:

Let suppose that fuel does not represent a significant amount in the rocket, so that rocket can be considered a constant mass system and can be used the supposition of constant acceleration both for climbing and descending.

a) The maximum altitude reached occurs when final speed is zero, the kinematic formula for the position of the rocket is:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot a \cdot t^{2}$

Where:

- Final height, measured in meters.

$y_{o}$ - Initial height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

- Time, measured in seconds.

- Acceleration, measured in meters per square second.

Aditionally, the speed as a function of position can be found by using the following kinematic expression:

$v^{2} = v_{o}^{2} + 2\cdot a \cdot (y-y_{o})$

Where is the final speed, measured in meters per second.

There are two situations that shall be analyzed prior determining the maximum altitude of rocket:

Climbing ( $a = 4\,\frac{m}{s^{2}}$ )

If first expression is used and $y_{o} = 0\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 5\,s$ , then the final height during the climbing stage is:

$y = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (5\,s) + \frac{1}{2}\cdot \left(4\,\frac{m}{s^{2}} \right)\cdot (5\,s)^{2}$

$y = 50\,m$

The speed of the rocket at the end of the stage is:

$v = 0\,\frac{m}{s} + \left(4\,\frac{m}{s^{2}} \right)\cdot (5\,s)$

$v = 20\,\frac{m}{s}$

Free-fall ( $a = -9.807\,\frac{m}{s^{2}}$ )

If second expression is used and $v_{o} = 20\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $y_{o} = 50\,m$ , the maximum altitude reached is determined after some algebraic handling:

$v^{2} = v_{o}^{2} + 2\cdot a \cdot (y-y_{o})$

$y-y_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$y = y_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$y = 50\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$y = 70.394\,m$

The rocket reaches a maximum height of 70.394 meters.

b) The time spent by the rocket during the free-fall climbing is found by deriving the position formula in time and replacing respective variables:

$v = v_{o} + a\cdot t$

$t = \frac{v-v_{o}}{a}$

$t = \frac{0\,\frac{m}{s}-20\,\frac{m}{s}}{-9.807\,\frac{m}{s^{2}} }$

$t = 2.040\,s$

Now, the time needed for the rocket to strikes the Earth from maximum altitude is derived from kinematic equation of position. That is to say:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot a \cdot t^{2}$

Given that $y_{o} = 70.394\,m$ , $y = 0\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $a = - 9.807\,\frac{m}{s^{2}}$ , the following second order polynomial is obtained:

$-4.904\cdot t^{2} + 70.394 = 0$ , where is measured in seconds.

The solution of the polynomial is:

$t = \sqrt{\frac{70.394}{4.904} }$

$t = 3.789\,s$

The total time elapsed from takeoff until impact is the sum of climbing, free-fall climbing and descent times:

$t_{T} = 5\,s + 2.040\,s + 3.789\,s$

$t_{T} = 10.829\,s$

The total time elapsed from takeoff until impact is 10.829 seconds.

Answer from: Quest

Lx bxh

Answer from: Quest

we need to know the time interval during which the force acts. impulse of force = momentum change

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