(i) Show that the maximum height reached by the first stage of the rocket is about 170 m.

(ii) On reaching its maximum height, the first stage of the rocket falls away and the second stage fires so that the rocket acquires a constant horizontal velocity of 56 m s^–1. Calculate the velocity at the instant when the second stage of the rocket returns to the surface of the Earth. Ignore air resistance.

A full-scale version of the rocket reaches a height of 260km when the first stage falls away. Using the data below, calculate the speed at which the second stage of the rocket will orbit the Earth at a height of 260km.

Mass of Earth = 6.0×10²⁴ kg
Radius of Earth = 6.4×10⁶ m

(i) attempt at area under graph;
appropriate triangle 175 m;
a comment about missing area making answer a little less / OWTTE;

(ii) \(t = \sqrt {\frac{{2 \times 170}}{{9.81}}} \left( { = 5.89{\rm{s}}} \right)\);
u=57.8(ms^–1) or u²=3340m²s⁻²;
speed ( \( = \sqrt {\left( {{{57.8}^2} + {{56}^2}} \right)}  = 80.4{\rm{m}}{{\rm{s}}^{ - 1}}\);
46° to horizontal;

\(\frac{{GMm}}{{{r^2}}} = \frac{{m{v^2}}}{r}\);
\(v = \sqrt {\frac{{GM}}{r}} \);
7.75×10³ms⁻¹;