1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DiKsa [7]
3 years ago
9

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

ng its initial line of motion with an acceleration of 32.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by the rocket. 102 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m (b) Find its total time of flight. s (c) Find its horizontal range. 186 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m
Physics
1 answer:
Serggg [28]3 years ago
5 0

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

You might be interested in
Which of these describe life sustaining functions that all organisms and cells perform\
allochka39001 [22]

Answer:

what are my options?

Explanation:

5 0
2 years ago
A 1kg object loses 20 j of gpe as it falls how far does it fall
jarptica [38.1K]
Gravitational potential energy (GPE) is energy within an object with respect to its height. The equation of gpe is:

GPE = mgh

where:
m = mass of object
g = gravitational constant, 9.81 m/s^2
h = height

For a given gpe of 20 J, the height can be calculated as follows:
*Remember to be consistent with units. Since Joule is derived from SI units, make sure to use SI units for the other variables

GPE = 20 J = 1 (9.81) (h)
h = 2.04 meters
4 0
3 years ago
A satellite is in a circular orbit around the Earth at an altitude of 3.18x10 m. Find the period and th orbital speed of the sat
Leona [35]

Answer:

112.17 m/s

56.427 years

Explanation:

h = 3.18 x 10^10 m

R = 6.4 x 10^6 m

r = R + h = 3.18064 x 10^10 m

M = 6 x 10^24 kg

The formula for the orbital velocity is given by

v = \sqrt{\frac{G M }{r}}

v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24}  }{3.18064\times 10^{10}}}

v = 112.17 m/s

Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17

T = 0.178 x 10^10 s

T = 56.427 years

8 0
3 years ago
I will give brainliest) According to Newton's second law of motion, when an object is acted on by an unbalanced force, how will
evablogger [386]
Newton's<span> first </span>law of motion<span> has been frequently stated throughout this lesson. An</span>object<span> at rest stays at rest and an </span>object<span> in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless </span>acted<span> upon by an </span>unbalanced force<span>.</span>
4 0
3 years ago
Read 2 more answers
The slope of the line tangent to the curve on a position-time graph at a specific time is the
Rudik [331]

Answer:

I do I make a brinliest can you please can me

7 0
3 years ago
Other questions:
  • Moving your finger across a textured surface can produce vibrations that are interpreted as texture. these vibrations are define
    6·1 answer
  • If an acid is combined with a base of equal strength, the result will most likely be
    9·1 answer
  • Two technicians are discussing disc brake rotor service. Technician A says that the inner races (bearing cups) should be removed
    15·1 answer
  • 12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L
    5·1 answer
  • Which vector best represents the net force acting on +3C charge in the diagram?
    10·2 answers
  • * An 10 kg box is sitting on a fairly smooth surface. Friction is still present. The coefficient of static friction
    7·1 answer
  • What are the 4 components that ybu need to see?
    12·1 answer
  • 15 POINTS PLZ HELP IM ON A TIMER! IM ON EDGE EXAM
    14·2 answers
  • A car 4m long moving at a velocity of 25m/s was beside a lorry 20m long with a velocity 19m/s at t=0. The distance between them
    6·1 answer
  • 3. Assume that a sample decays with a given half-life. Will you be able to observe when the sample has completely decayed? Why o
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!