Question

A Diwali rocket ascends with a net upward acceleration \[20m{s^{ - 2}}\] from rest at \[t = 0\] seconds. The fuel gets exhausted at \[t = 10\] seconds and continues to move up. Find the maximum height attained by it.

Answer 1

Hint:When the rocket is accelerated, it reaches a height of \[1000\,m\] and continues to move upward due to some velocity. At its peak point, the rocket will have a velocity \[{v_2}\] of \[0\,m/s\].

Formula used:
Distance formula
$S=ut+\dfrac{1}{2}at^2$
Where, \[S = \] Distance
\[u = \] Velocity
\[t = \] Time
\[a = \] Acceleration
Initial velocity, \[v{\rm{ }} = {\rm{ }}u{\rm{ }} + {\rm{ }}at\]
Where, \[u = \] Velocity
\[t = \] Time
\[a = \] Acceleration
External velocity, $V^2=v^2+2gs$
\[v = \] Initial velocity
\[s = \] distance
\[g = \]gravitational acceleration

Complete step by step solution:
Velocity, \[u = 10m/s\]
Net acceleration, $a=20\,m/s^2$
The gravity acceleration, $g=-10\,m/s^2$
Time, \[T{\rm{ }} = {\rm{ }}10{\rm{ }}s\]
We know that $S=ut+\dfrac{1}{2}at^2$
Substitute all the values and we get
$S=0(10)+\dfrac{1}{2}(20)(10)^2$
Simplifying and we get
\[S{\rm{ }} = {\rm{ }}1000{\rm{ }}m\]
Let's say the rocket had a velocity of \[{V_1}\] when it ascended \[1000m\].
\[{V_1}{\rm{ }} = {\rm{ }}u{\rm{ }} + {\rm{ }}at\]
Substituting the values and we get
\[ \Rightarrow {V_1}{\rm{ }} = {\rm{ }}0{\rm{ }} + {\rm{ }}20(10)\]
Simplifying and we get
\[ \Rightarrow {V_1}{\rm{ }} = {\rm{ }}200{\rm{ }}m/s\]
when the rocket was at a height of \[1000m\], it was moving upward with a velocity of \[ \Rightarrow {V_1}{\rm{ }} = {\rm{ }}200{\rm{ }}m/s\].
At its peak point, the rocket will have a velocity \[{V_2}\] of \[0m/s\] .

Substituting the values and we get
\[ \Rightarrow 0{\rm{ }} = {\rm{ }}200(200){\rm{ }} - {\rm{ }}2(10)s\]
Simplifying and we get
\[ \Rightarrow s{\rm{ }} = {\rm{ }}2000m\]
Total height attained by rocket \[ = 1000{\rm{ }} + {\rm{ }}2000{\rm{ }}= {\rm{ }}3000{\rm{ }}m\]

Therefore, the maximum height attained by it is 3000m.

NoteHere the value of acceleration due to gravity is taken as \[10\,m/s^2\] but if mentioned in the question acceleration due to gravity should be taken as \[9.8\,m/{s^2}\]. Acceleration is defined as the rate of change of velocity with respect to time. Since acceleration has both a magnitude and a direction, it is a vector quantity. Additionally, it is the first derivative of velocity with respect to time or the second derivative of position with respect to time.