Help with finding Velocity...........?

Question

A rocket moves striaght upward, starting from rest with an acceleration of 29.4m/s^2. It runs out of fuel at the end of 4.00seconds and continues to coast upward, reaching a maximum height before falling back to Earth.

(a) Find the velocity the instant before the rocket crashes on the ground.





Not sure how to do this. Can anyone help walk me through it??
Thanks <3<3<3

Is it 157m/s or 136m/s? Im confused

Answer 1

1) find velocity at end of thrust. (Vf = Vo + at)
2) find height attained during thrust phase (d = .5*(Vo+Vf)*t)
3) find max height during coasting ( Vf² = Vo² + 2ad where Vf = 0 and Vo = Vf from part 1)
4) find final velocity as rocket accelerates back to earth. (Vf² = Vo² + 2ad where Vo = 0, a = 9.8 m/s², and d was calculated in 2 and 3


solution......

1)

Vf = Vo + at = 0 + 29.4 m/s² x 4.00 s = 117.6 m/s

2)

d = .5 x (0 + 117.6 m/s) x 4.00 s = 235.2 m

3)

Vf² = Vo² + 2ad
d = (Vf² - Vo²) / 2a = ( 0² - (117.6 m/s)²) / (2 x -9.8 m/s²)
= 705.6 m

4)

first, the height of the rocket at it's peak is 235.2 m + 705.6 m = 940.8 m

from
Vf² = Vo² + 2ad = 0 + 2 x 9.8 m/s² x 940.8 m = 18440 m²/s²

so that Vf = 136 m/s

(three sig figs right?)


*******edward*******

thanks for the thumbs down dude.

you maybe want to fix your answer....

you have d = a x t + 1/2 at^2 = 29.4 m/s^2 x 4.00 s + ...

problem is.....

29.4 m/s^2 x 4 s = 117.6 m/s not 117.6 m

what you should have instead is d = Vot + 1/2 at^2 for both parts. since Vo = 0, d = 1/2 at^2 for the first section too. ie..

.5 x 29.4 m/s^2 x (4s)^2 = 235.2 m + ......

in which case our answers agree....


another update for edward.....

dude....

d = 1/2 x a x t1^2 + 1/2 a^2/g x t1^2
= .5 x (29.4 m/s^2) x (4s)^2 + .5 x (29.6 m/s^2)^2 x (4s)^2 / (9.8 m/s^2)

= 235.2 m + 705.6 m²s²/(s^4 x m / s^2)
= 235.2 m + 705.6 m²s²/(s^2 x m) = 235.2 m + 705.6 m
= 940.8 m

then.....

V = sqrt (2 x g x h) = sqrt ( 2 x 9.8 m/s^2 x 940.8 m) = 136 m/s

same answer as mine


**********sheff**********
don't be confused. the correct answer is 136 m/s. it works out to that answer both methods. edward goofed up his math. he'll correct it when he realizes it.

Answer 2

Have you lost your velocity?

When it reaches it's hiest point it will have potential energy Pe and then it will hit the groung with kinetic energy Ke

Pe=Kef then
mgh=0.5 m V^2
V^2=2gh
and
V = sqrt(2 gh)
so what is h

h=at1^2 + 0.5gt2
t2= V/g = a t1/g =(a/g) t1 finally
h=a t1^2 + 0.5 g (a/g)^2 t1^2
h=(a + .5 g(a/g)^2)t1^2
h=[29.4 + 0.5 x 9.8 (29.4/9.8)^2] (4)^2=
h=1180m


since V = sqrt(2 gh) we have

V = sqrt(2 x 9.8 x 1180)=152 m/s

OR simply

V=gt=g(t1+t2)
V=g(t1 + (a/g)t1)
V=g(1+a/g) t1
V=9.8( 1 + 29.4/9.8)4
V=157 m/s

curious results