This question is about the motion of a firework rocket.

A firework rocket is fired vertically upwards from the ground. It accelerates uniformly from rest
with an acceleration of 8.0 for 5.0 s after which time the fuel of the rocket has all been used. 2 ms−
(a) (i) Sketch below a graph to show how the velocity of the rocket changes with time from
the moment it leaves the ground until the moment that it returns to the ground. Mark
on your sketch the time at which the fuel has run out, the time at which the rocket 1
t 2t
reaches its maximum height and the time at which it reaches the ground. 3t
(Note that you are not expected to give any quantitative values of velocity and time and
air resistance can be ignored.)
(ii) Comment on the area(s) under the graph that you have drawn.
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In the following calculations you may ignore any effects of air resistance and take the acceleration
due to gravity, g 10 m s . −2 =
(b) Calculate the
(i) speed of the rocket when the fuel runs out.
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(ii) height that the rocket reaches when the fuel has just run out.
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(iii) maximum height reached by the rocket.
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(iv) time it takes the rocket to reach its maximum height.
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(v) time that it takes to fall from its maximum height to the ground.
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(c) (i) On the axes below sketch graphs to show how the gravitational potential energy and the
kinetic energy of the rocket varies as it moves from the ground to its maximum height.
(Note that this is only a sketch graph; you do not need to add any numerical values.)
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(ii) State one assumption, other than ignoring air resistance, that you have made in
sketching the above graph.
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(d) The rocket plus fuel initially have a mass of 0.16 kg. If the initial mass of the fuel is 0.02 kg,
calculate the maximum kinetic energy of the rocket when all the fuel has been used.
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(e) Calculate the power delivered to the rocket by the rocket fuel. [2]
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(f) Describe two consequences on the motion of the rocket as a result of air resistance acting on
the rocket.
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Sorry its physics

Surely you have some ideas on some of the parts.

It's all based on h(t) = h0 + v0t + at^2
KE = 1/2 mv^2

So, how far do you get?

oops ... 1/2 at^2