Question

A projectile is fired vertically from earth's surface with an initial speed of 7.3 km/s. neglecting air drag, how far above the surface of earth will it go?

Answer 1

We can solve the problem by using conservation of energy. 

In fact, initially the projectile has only kinetic energy, which is given by
$K= \frac{1}{2}mv^2$
where m is the projectile's mass while $v=7.3 km/s=7300 m/s$ 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
$U=mgh$
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:
$\frac{1}{2}mv^2=mgh$
$h= \frac{v^2}{2g}= \frac{(7300 m/s)^2}{2 \cdot 9.81 m/s^2}=2.72 \cdot 10^6 m = 2720 km$