We can solve the problem by using conservation of energy.
In fact, initially the projectile has only kinetic energy, which is given by
[tex]K= \frac{1}{2}mv^2 [/tex]
where m is the projectile's mass while [tex]v=7.3 km/s=7300 m/s[/tex] is its initial velocity.
At the point of maximum height, the speed of the projectile is zero, so it only has gravitational potential energy which is equal to
[tex]U=mgh[/tex]
where g is the gravitational acceleration and h is the maximum height of the projectile.
Since the energy must be conserved, we can equalize K and U to find the value of h:
[tex] \frac{1}{2}mv^2=mgh [/tex]
[tex]h= \frac{v^2}{2g}= \frac{(7300 m/s)^2}{2 \cdot 9.81 m/s^2}=2.72 \cdot 10^6 m = 2720 km [/tex]
