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Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rocket ship accelerations upwards at 2g), but runs out of fuel after 100 seconds, after which point it stops accelerating upward. At this point, the rocket begins accelerating downwards with a magnitude of g. Assume that the gravitational pull of the Earth on the rocket doesn't change with altitude.

a) How high above the surface of the Earth does the rocket travel before it stops accelerating?
b) How fast is the rocket going when it stops accelerating?
c) How high does the rocket get before it begins to fall back to Earth?
d) How long after launch does the rocket strike the Earth again?
e) What is the average velocity of the rocket between when it leaves the Earth and when it strikes the Earth again?

Respuesta :

Answer:

a) [tex]s_a=98100\ m[/tex] is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) [tex]v_a=1962\ m.s^{-1}[/tex] is speed of the rocket going when it stops accelerating.

c) [tex]H=294300\ m[/tex]

d) [tex]t_T=544.95\ s[/tex]

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, [tex]a=2g=2\times 9.81=19.62\ m.s^{-2}[/tex]

time for which the rocket accelerates, [tex]t_a=100\ s[/tex]

For the course of upward acceleration:

using eq. of motion,

[tex]s_a=ut+\frac{1}{2}at_a^2[/tex]

where:

[tex]u=[/tex] initial velocity of the rocket at the launch [tex]=0[/tex]

[tex]s_a=[/tex] height the rocket travels just before its fuel finishes off

so,

[tex]s_a=0+\frac{1}{2}\times 19.62\times 100^2[/tex]

a) [tex]s_a=98100\ m[/tex] is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

Now the velocity of the rocket just after the fuel is finished:

[tex]v_a=u+at_a[/tex]

[tex]v_a=0+19.62\times 100[/tex]

b) [tex]v_a=1962\ m.s^{-1}[/tex] is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

[tex]v^2=v_a^2-2gh[/tex]

where:

[tex]g=[/tex] acceleration due to gravity

[tex]v=[/tex] final velocity of the rocket at the top height

[tex]0^2=1962^2-2\times 9.81\times h[/tex]

[tex]h=196200\ m[/tex]

c) So the total height at which the rocket gets:

[tex]H=h+s[/tex]

[tex]H=196200+98100[/tex]

[tex]H=294300\ m[/tex]

d)

Time taken by the rocket to reach the top height after the fuel is over:

[tex]v=v_a+g.t[/tex]

[tex]0=1962-9.81t[/tex]

[tex]t=200\ s[/tex]

Now the time taken to fall from the total height:

[tex]H=v.t'+\frac{1}{2}\times gt'^2[/tex]

[tex]294300=0+0.5\times 9.81\times t'^2[/tex]

[tex]t'=244.95\ s[/tex]

Hence the total time taken by the rocket to strike back on the earth:

[tex]t_T=t_a+t+t'[/tex]

[tex]t_T=100+200+244.95[/tex]

[tex]t_T=544.95\ s[/tex]

e)

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

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