Question

A rocket of $3\times {10}^{6}kg$ mass takes off from a launching pad and acquires a vertical velocity of $1km{s}^{-1}$at an altitude of $25km$.Calculate

(i) $PE$

(ii) $KE$

Take $\left(g=10m/s^2\right)$

Answer 1

Step 1: Given data:

Mass $m=3\times {10}^{6}kg$

$h=25km$

We know that $1km=1000m$,therefore

$$\begin{gathered} h=25km \\ \Rightarrow 25000m \end{gathered}$$

Velocity $v=1km{s}^{-1}$

We know that $1km=1000m$, therefore

$v=1000m{s}^{-1}$

$g=10m/s^2$

Step 2: Calculation of PE:

We know the formula for potential energy,

$PE=mgh$, where $m$ is mass, $PE$ is potential energy, $g$ is acceleration due to gravity and $h$ is height.

$PE=3\times {10}^6\times 10\times 25000$

$=75000\times {10}^6\times 10$

$=75\times {10}^3\times {10}^6\times 10$

$=75\times {10}^{10}$

$=7.5\times {10}^{11}J$

Hence for a rocket of $3\times {10}^{6}kg$ mass takes off from a launching pad and acquires a vertical velocity of $1km{s}^{-1}$ at an altitude of $25km$ the potential energy is $7.5\times {10}^{11}J$.

Step 3: Calculation of KE:

We know the formula for kinetic energy,

$KE=\frac{1}{2}m{v}^2$, where $m$ is mass, $KE$ is kinetic energy and $V$ is velocity.

$KE=\frac{1}{2}\times 3\times {10}^6\times {\left(1000\right)}^2$

$=1.5\times {10}^6\times {10}^6$

$=1.5\times {10}^{12}J$

Hence for a rocket of $3\times {10}^{6}kg$mass takes off from a launching pad and acquires a vertical velocity of $1km{s}^{-1}$at an altitude of $25km$ the kinetic energy is $1.5\times {10}^{12}J$.