A rocket rises vertically, from rest, with an acceleration of 3.2 m/s^2 until it

Question

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s^2 until it runs out of fuel at an altitude of 775 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 it strike the Earth? (f) How long (total) is it in the air?

Explanation / Answer

a = 3.2 m/s^2

a) from the equation v^2-u^2 = 2ax

v = sqrt(2*3.2*775) = 70.43 m/s

b) t = v/a = 70.43/3.2 = 22.01 sec

c) T = 70.43/9.8 = 7.2 sec , 22.01 s to reach 775 m

s = 70.43*7.2-1/2*9.8*7.2^2 = 506.2-254.02 = 252.18 m

h = 252.18+775 = 1027.18 m

d) t' = 7.2+22.01 = 29.21 s

e) 1/2*9.8*t'^2 = 1027.18

t' = 14.48 s

v' = 9.8*14.48 = 141.9 m/s

f) t tot = 14.48+29.21 = 43.69 s

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