A rocket rises vertically, from rest, with an acceleration of 2.6 m/s2 until it

Question

A rocket rises vertically, from rest, with an acceleration of 2.6 m/s2 until it runs out of fuel at an altitude of 1000 m. After this point, its acceleration is that of gravity, downward. (a) What is the velocity of the rocket when it runs out of fuel? (Assume up is positive.) m/s

(b) How long does it take to reach this point? s

(c) What maximum altitude does the rocket reach? m

(d) How much time (total) does it take to reach maximum altitude? s

(e) With what velocity does the rocket strike the Earth? (Assume up is positive.) m/s

(f) How long (total) is it in the air? s

Explanation / Answer

a)

V^2 - u^2 = 2*a*s

V^2 - 0 = 2*2.6*1000

V = 72.11 m/s

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b) s = ut + 0.5*a*t1^2

1000 = 0.5*2.6*t1^2

t1 = 27.73 s
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c)

h = v^2/2g = 265.3

Hmax = s + h = 1265.3 m

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d)

s = ut + 0.5*a*t1^2

1000 = 0.5*2.6*t1^2

t1 = 27.73 s

after fuel runs out it reaches max height at max height Vf = 0

V = 0 - gt2

t2 = v/g = 7.36 s

T = t1+t2 = 35.09 s

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e) from max height

initial velocity = 0

y = 1265.3 m

Vf = ?

Vf^2 - 0 = 2*g*y

Vf^2 = 2*9.8*1265.3

Vf = -157.5 m/s

f)

t3 = vf - 0/g

t3 = 16.07 s

Total time in air = t1+t2+t3 = 51.16 s

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