Question

An object is projected upward from the surface of the earth with an initial speed of 3.7 km/s. find the maximum height it reaches.

Answer 1

The kinetic energy of the object just after it starts its motion from the Earth surface is
$K= \frac{1}{2}mv^2$
where m is the object mass and $v=3.7 km/s=3700 m/s$ its initial speed.
When it reaches its maximum height, the object speed is zero and all its kinetic energy converted into gravitational potential energy, which is
$U=mgh$
where $g=9.81 m/s^2$ and h is the maximum height reached by the object.

Since the energy of the object must be conserved, K=U, therefore we can write
$\frac{1}{2}mv^2 =mgh$
and we can solve to find h, the maximum height:
$h= \frac{v^2}{2g}= \frac{(3700 m/s)^2}{2\cdot 9.81 m/s^2}= 6.98 \cdot 10^5 m=698 km$