Question: A projectile is fired vertically from Earth’s surface with an initial speed of $\mathbf{10}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{s}$. Neglecting air drag, how far above the surface of Earth will it go?

The initial speed of a projectile is

$$\begin{gathered} v_1=10\frac{\text{km}}{\text{s}} \\ =10,000\text{m/s}. \end{gathered}$$

According to the rule of conservation of energy, energy can only be transformed from one form of energy to another and cannot be created or destroyed.

Using the law of conservation of energy, we can find the distance of the projectile from earth’s surface. From this, we can find the maximum height achieved by the projectile.

Formula:

$\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

Where, m is the mass of object, M is the mass of earth, v1 is the initial speed of a projectile and v is the final speed of a projectile

We have the equation for conservation of energy as,

$$\begin{gathered} K_1+U_1=K_2+U_2 \\ \frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}^2-\frac{GMm}{r} \end{gathered}$$

Now, substituting the known values,

$$\begin{gathered} G=6.67\times {10}^{-11}\frac{m^3}{kg{s}^2} \\ M=5.98\times {10}^{24}kg \\ {r}_1=6.37\times {10}^6\text{m} \\ v=0m/s \end{gathered}$$

We can solve for r as,

$$\begin{gathered} \frac{1}{2}{\left(10,000\text{m/s}\right)}^2-\frac{\left(6.67\times {10}^{-11}\frac{m^3}{kg{s}^2}\right)\times \left(5.98\times {10}^{24}kg\right)}{6.37\times {10}^6\text{m}}=-\frac{\left(6.67\times {10}^{-11}\frac{m^3}{kg{s}^2}\right)\times \left(5.98\times {10}^{24}kg\right)}{r} \\ r=3.2\times {10}^{7}m \end{gathered}$$

So, the maximum height is given by

$$\begin{gathered} h=r-r_1 \\ =2.5\times {10}^{7}m \end{gathered}$$

Therefore, the maximum height achieved by the projectile is $h=2.5\times {10}^{7}m$.