Hello, I need to hand this in tomorrow. It's within a Calculus paper.

What I'm doing is wrong as I need to work out the answer to part b before I can answer part a. Plus this is a calculus question and I'm not sure how that fits in with my attempt at an answer. There must be an easier less complicated way.A Rocket is modelled by a particle which moves along a vertical line. From launch, the rocket rises until its motor cuts out after 14 seconds. At this time it has reached a height of 480 metres above the launch pad and attained an upward velocity of $\displaystyle 70ms^{-1}$. 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 origin at the launch pad. Take $\displaystyle t=0$ 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 (form launch) does the rocket take to reach this maximum height?

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

Formula for deceleration;

b)

$\displaystyle deceleration = \frac{(Inital Velocity - final Velocity)} {time}$

We know at the point of it's maximum;

$\displaystyle -10ms^{-2} = \frac{(70ms^{-1} - 0)} t $

$\displaystyle t = \frac{70ms^{-1}} {-10ms^{-2}}$

a)

$\displaystyle average velocity = \frac {distance} {time}$

from part (b) we know the know the time $\displaystyle (t)$. Therefore;

$\displaystyle 70ms^{-1}+10ms^{-2} = \frac {d} {t}$

$\displaystyle d = \frac{70ms^{-1} +10ms^{-2}} {t}$

c)

$\displaystyle 14 seconds + t + x $

$\displaystyle acceleration = \frac {Final Velocity - Initial Velocity} {time}$

$\displaystyle -10ms^{-2} = \frac {Final Velocity - 70ms^{-1}} {x}$

$\displaystyle x = \frac {Final Velocity - 70ms^{-1}} {-10ms^{-2}}$