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aniis · May 16, 2009, 05:43 AM

A firework rocket starts from rest at ground level and moves vertically. In the first 3 s of its motion, the rocket rises 27m. The rocket is modelled as a particle moving with constant acceleration 6 m/s2

a) Find the speed of the rocket 3 s after it has left the ground.

After 3 s, the rocket burns out. The motion of the rocket is now modelled as that of a particle moving freely under gravity.

b) Find the height of the rocket above the groud 5 s after it has left the ground.
:eek:

Perito · May 16, 2009, 05:47 AM

You assume a constant force on the rocket until the rocket motor burns out. Since the initial velocity is 0, this is.

$X=\frac {1}{2} AT^2$

After that, it is in free fall, it's the same equation, but you have the acceleration (deceleration) due to gravity.

$X = V_oT + \frac 12 GT^2$

aniis · May 16, 2009, 05:56 AM

s=ut+ 1/2 at2
that is what I did to find out the acceleration.
What do I do to find out the speed? I only know it should give me 18m/s

Perito · May 16, 2009, 06:13 AM

s=ut+ 1/2 at2
that is what I did to find out the acceleration.
What do I do to find out the speed? I only know it should give me 18m/s

You know the acceleration: "constant acceleration 6 m/s^2. You solve for the distance.

Unknown008 · May 16, 2009, 10:02 AM

Originally Posted by aniis

s=ut+ 1/2 at2
that is what I did to find out the acceleration.
What do I do to find out the speed? I only know it should give me 18m/s

You can also use

$v^2=u^2 +2as$

v is the final velocity (? m/s)
u is the initial velocity (0 m/s)
a is the acceleration (6 m/s^2)
s is the distance.(27 m)

Or, you use

$v = u + at$

You'll have the same answer in each case.

For b, you use

$s=ut+ \frac{1}{2}at^2$

s is the unknown distance (? m)
u the initial velocity (0 m/s)
t time (5 s)
a acceleration (6 m/s)

Justto make things clearer.

Hope that helped! :)