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:

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



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

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

Quote:

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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

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

Quote:

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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

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You can also use



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



You'll have the same answer in each case.

For b, you use



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! :)