Physics, 20.09.2020 15:01 annetteaudc

A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2 . It runs out of fuel at the end of 4.00s and continues to coast upward, reaching a maximum height before falling back to earth. (a) Find the rocket’s velocity and position at the end of 4.00s. (b) Find the maximum

Answers: 3

Show answers

Answers

a) The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively, b) The maximum height reached by the rocket is 940.296 meters, c) The rocket crashes on the ground at a velocity of -96.030 meters per second.

Explanation:

The complete statement is:

A rocket moves straight upward , starting from rest with an acceleration of 29.4 m/s2. it runs out of fuel at the end of 4.00 s and countinues to coast upward , reaching a maximum height before falling back to earth . a) find the rocket's velocity and position at the end of 4.00 s . b) Find the maximum height the rocket reaches. c) find the velocity on the instant before the rocket crashes on the ground.

a) The rocket accelerates uniformly, whose equations of motion are:

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

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

Where:

- Final position, measured in meters.

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

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

- Time, measured in second.

- Acceleration, measured in meters per square second.

- Final velocity, measured in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o} = 0\,\frac{m}{s}$ , $t = 4\,s$ and $a = 29.4\,\frac{m}{s^{2}}$ , the velocity and position of the rocket are, respectively:

$y = 0\,m+\left(0\,\frac{m}{s} \right)\cdot (4\,s)+\frac{1}{2}\cdot \left(29.4\,\frac{m}{s^{2}} \right) \cdot (4\,s)^{2}$

$y = 235.2\,m$

$v = 0\,\frac{m}{s} + \left(29.4\,\frac{m}{s^{2}}\right)\cdot (4\,s)$

$v = 117.6\,\frac{m}{s}$

The velocity and position of the rocket at the end of 4 seconds are 117.6 meters per second and 235.2 meters, respectively.

b) Now, the rocket experiments a free-fall motion. The maximum height of the rocket is obtained by equalizing the equation of velocity to zero and evaluating the equation of position later. That is:

$y = 235.2\,m+\left(117.6\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}$

$v = 117.6\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right)\cdot t$

Then,

$0 = 117.6 -9.807\cdot t$

$t = 11.991\,s$

The maximum height reached by the rocket is:

$y = 235.2+117.6\cdot (11.991)+\frac{1}{2}\cdot (-9.807)\cdot (11.991)^{2}$

$y = 940.296\,m$

The maximum height reached by the rocket is 940.296 meters.

c) The rocket experiments a free-fall motion and is accelerated by gravity until collision happens. The equations of motion below are presented:

$0\,m = 940.296\,m+\left(0\,\frac{m}{s} \right)\cdot t+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}$

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

From the equation of position we get the instant when rocket hits the ground, whose roots are found by Quadratic Formula:

$t_{1} \approx 9.792\,s$ and $t_{2} \approx -9.792\,s$

Only the first root is physically reasonable.

By the second equation we calculate the final velocity:

$v = 0-9.807\cdot (9.792)$

$v = -96.030\,\frac{m}{s}$

The rocket crashes on the ground at a velocity of -96.030 meters per second.

Answer from: Quest

That would be kinetic energy

Answer from: Quest

the maxium height is 34m/s

Another question on Physics

Physics, 22.06.2019 04:50

Unpolarized light whose intensity is 1.19 w/ is incident on a polarizer. (a) what is the intensity of the light leaving the polarizer? (b) if the analyzer is set at an angle of = 41.0∘ with respect to the polarizer, what is the intensity of the light that reaches the photocell?

Answers: 1

Answer

Physics, 22.06.2019 22:00

What would people living along the coast in south florida do if there was a hurricane warning? move to locations away from the water flock along coasts to watch the natural phenomenon buy instruments to predict the exact location of the hurricane measure water levels to know the exact time of the hurricane

Answers: 3

Answer

Physics, 22.06.2019 22:30

Isobars are lines of constant or equal pressure on a weather map. they can be used to find areas of low or high pressure over a broad area, and they can tell us how intense the system may be. on weather maps, you may have noticed areas that have a large l or h over a region with lines circling around them. the l stands for low pressure and h stands for high pressure. the lines circling them are isobars. check out the pressure system and isobars over georgia and florida. what is the present weather in this area ? a) clear and sunny b) cold and cloudy c) changing weather d) lots of precipitation

Answers: 3

Answer

Physics, 22.06.2019 23:50

What would be the radius of the earth if it had its actual mass but had the density of nuclei? express your answer to two significant figures and include the appropriate units.

Answers: 2

Answer

You know the right answer?

A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/s2 . It runs out...

Questions