Physics, 11.02.2021 21:00 7obadiah

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreciable air resistance. When it has reached a height of 595 m, its engines suddenly fail so that the only force acting on it is now gravity. A) What is the maximum height this rocket will reach above the launch pad?
B) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?

Answers: 2

Show answers

Answers

Answer from: elizabethprasad2

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

- Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

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

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch (), measured in seconds, is:

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

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

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

Where:

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

- Final speed, measured in meters per second.

- Gravitational acceleration, measured in meters per square second.

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

- Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

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

- Final height, measured in meters.

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

- Gravitational acceleration, measured in meters per square second.

- Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

Answer from: Quest

distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

slope of a graph = change in value of y -axis / change in values of x -axis

comparing both the equations

change in value of y -axis = change in position

change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.

Answer from: Quest

Ineed the problem dude

Another question on Physics

Physics, 22.06.2019 14:30

A10nc charge sits at a point in space where the magnitude of the electric field is 1500 n/c. what will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? assume the system is big enough to consider the charges as small test charges.

Answers: 1

Answer

Physics, 22.06.2019 16:40

The astronauts on the space shuttle flights experienced an acceleration of 29 m/s2 (about 3 "g's") during lift-off. what upward force must the astronaut's seat apply to the astronaut in order to cause this acceleration? assume the astronaut's mass is 70 kg, and compute this force when the acceleration is near the earth's surface so their weight equals latex: mg m g .

Answers: 3

Answer

Physics, 22.06.2019 17:40

Aball is thrown horizontally from a cliff at 8m/s .in what direction is the ball moving 2s later?

Answers: 1

Answer

Physics, 22.06.2019 23:30

Using a caesar cypher with an offset of three characters (a -> d, b -> e, z -> c), what would be the correct cypher text for the plain text "attack at dawn"?

Answers: 1

Answer

You know the right answer?

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2....

Questions