Modelling rocket motion

[val1]

Please can you help with this one?

A rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 18 seconds. At this time it has reached a height of 540 metres above the launch pad and attained an upward velocity of \(\displaystyle \ 60ms^{ - 1} \\). From this time on, the rocket has constant upward acceleration \(\displaystyle \ - 10ms^{ - 2} \\) due to the effect of gravity alone.

Choose the s-axis (for the position of the particle that represents the rocket) to point upwards, with the origin at the launch pad. Take t=0 to be the time when the rocket motor cuts out.

a) What is the maximum height (above the launch pad) reached by the rocket?

b) How long (from launch) does the rocket take to reach this maximum height?

c) After how long (from launch) does the rocket crash onto the launch pad?


For part a) I used the formula \(\displaystyle \ v^2 - 2as = v_0 ^2 - 2as_0 \\)

(where \(\displaystyle \ v_0 \\) is the initial velocity at the point of origin and \(\displaystyle \ s_0 \\) is the initial distance from the origin)

Putting \(\displaystyle \ v_0 = 60 \\) , \(\displaystyle \ s_0 = 540 \\) , \(\displaystyle \
a = - 10 \\) and v=0 (max. height is reached when velocity is zero), I got


\(\displaystyle \ \frac{{60^2 + 2 \times 10 \times 540}}{{ - 2 \times - 10}} = 720 \\)

Is this is correct?

I have got completely stuck on the other two parts. I have to use the formula \(\displaystyle \s = \frac{1}{2}at^2 + v_0 t + s_0 \\) but am unsure what values to substitute to find the relevant times. I was offered some help with this question, using another method. I appreciate the help, but it didn't help me at all as I didn't understand the user's method. Also, I have to use the given formula, or a derivation of it, for this problem. I am very stuck.

Please can you help?

 

[morson]

The formulae you mentioned only work for constant acceleration. The rocket's
acceleration varies up to the 540m point above the launch pad, so it's from here
where we use the formulas. Letting the launchpad be the origin:

v = u + at, where u is the initial speed and a is the constant acceleration

Plug in u = 60, a = -10:

v = 60 - 10t, but v = ds/dt

By direct integration, s = 60t - 5t^2 + c

We know that when t = 0, s = 540:

540 = c ==> s = 60t - 5t^2 + 540

a) The maximum height is simply the vertex of the parabola. The vertex of a quadratic
y = ax^2 + bx + c is located at (-b/2a, -M/4a), where M is the discriminant. For this
particular quadratic s = 60t - 5t^2 + 540, b = 60, a = -5 and M = 14400, so the vertex
is therefore (6, 720), so this is a maximum height of 720 metres above the launchpad.

So yes, you were correct. Good work.

b) You could just use the method above or set v = 0 and solve for t. But remember, t = 6 means 6 seconds after the motor cuts out, so the rocket actually
reaches its maximum height 24 seconds from launch.

c) Set s = 0 and solve the quadratic equation.

60t - 5t^2 + 540 = 0

Hint: take out the 5 as a common factor first, and add on 18 to your valid solutions
as it asks "from launch."