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 1400 m. After this point, its acceleration is that of gravity, downward. (Assume that the positive direction is upward.) (a) What is the velocity of the rocket when it runs out of fuel? 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 it strike the Earth? m/s (f) How long (total) is it in the air?

Explanation / Answer

S=1/2 at^2 =0.5x3.4xt^2=1400 m t=28.69 sec . a) v=u+at u=0 so v= 97.57 m/sec ....Ans b)it take to reach this point s2= 28.69 sec .....Ans c) u^2=-2as ; v=0 when it stop so s=485.70 m so attutude is 1400+485.70 = 1885.70m .....Ans d)u=-at; 9.95 sec =t so total time t took is 38.64 sec ......Ans e)2as=v^2 going downword s=1885.70 m v=192.24 m/sec ........Ans f) v=at t=19.61 sec total time in air is 19.61+38.64 =58.25 sec ........Ans plsz rate me 1st ... i m correct

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