Question

A firework rocket takes off vertically with an acceleration of `20ms^-2` and maintains that acceleration for 5 seconds after which the rocket stops burning. Find (a), (b) and (c), below?

A firework rocket takes off vertically with an acceleration of `20ms^-2` and
maintains that acceleration for 5 seconds after which the rocket stops burning.

(a) What is the greatest height that the rocket eventually
reaches?

(b) How long does it take to reach its greatest height?

(c) How long does it take to fall back to the ground from its
greatest height?

Answer 1

a) 760m
b) 15.16s
c)12.4s

Explanation:

a)

First, we find the final speed and height of the rocket when it stops accelerating.

FInd the speed using `v=u+at`
u=0 because it takes off from rest

`v_1=0+20xx5`

`v_1=100 ms^-1`

And for the height in this time, `s=ut+1/2at^2`

`h_1=0+1/2xx20xx5^2`
`=250m`

Now, there is effectively a different journey. Here, the rocket starts with velocity of `100 ms^-1`, and has an acceleration of `-9.81ms^-2` (if we take up to be positive)

At its maximum height, the velocity of the rocket will be 0.

We can use `v^2=u^2+2as`

`0=100^2-2xx9.81xxs`

`19.62s=10000`
`s=509.68m`

Add our two heights to get the max height reached.

`h=250+510`
`=760m`

b)

The time for the first section is 5s, so we only need the second journey time.

We can use `v=u+at` here.

`0=100-9.81t`
`9.81t=100`
`t=10.19s`

So the total time is
`t_"tot"=5+10.19`
`=15.16"s"`

c)

Here, we are freefalling under gravity a distance of 760m with an initial speed of 0.

Use `s=ut+1/2at^2`

`760=0t+1/2xx9.81xxt^2`
We can take gravity as positive this time if we want to (which stops dealing with awkward negatives).

`760=4.905t^2`
`t^2=154.94`
`t=12.4"s"` (`t>0` so we can square root here)