A rocket rises vertically, from rest, with an acceleration of 3.4 m/s2 until it

Question

A rocket rises vertically, from rest, with an acceleration of 3.4 m/s2 until it runs out of fuel at an altitude of 1500 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?

B) How long does it take to reach this point?

C) What maximum altitude does the rocket reach?

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

E) With what velocity does the rocket strike the Earth?

F) How long (total) is it in the air?

Explanation / Answer

A) initial velocity is Vo = 0 m/sec

accelaration is a = 3.4 m/s^2

distance travelled is S = 1500 m

then using kinematic equations

V^2 -Vo^2 = 2*a*S

V^2 - 0^2 = 2*3.4*1500

V = sqrt(2*3.4*1500) = 101 m/sec

B) using V = Vo +(a*t)

101 = 0+(3.4*t)

t = 101/3.4 = 29.7 sec

C) maximum altitude reached is 1500 + distance travelled under gravitational force (S)

Hmax = 1500 + S

S = V^2/(2*g) = 101^2/(2*9.8) = 520.5 m

so Hmax = 1500+520.5 = 2020.5 m

D) t = t1+ t2

t1 = 29.7 sec

t2 = V/g = 101/9.8 = 10.3 sec

then t = 29.7 + 10.3 = 40 sec

E) V = sqrt(2*g*Hmax) = sqrt(2*9.8*2020.5) = 199 m/sec

F) T = t + td = 40 + td

using S = (Vo*t) + (0.5*a*td^2)

2020.5 = (0*t) + (0.5*9.8*td^2)

td = 20.3 sec

then T = t+ td = 40+20.3 =60.3 sec

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