Question

A rocket is fired upward from the earth’s surface such that it creates an acceleration of ${\text{19}}{\text{.6 m}}{{\text{s}}^{{\text{ - 2}}}}$. If after ${\text{5 sec}}{\text{.}}$, its engine is switched off, the maximum height of the rocket from earth’s surface would be
A. $980{\text{ m}}$
B. ${\text{735 m}}$
C. $490{\text{ m}}$
D. ${\text{245 m}}$

Answer 1

Hint:In this question, we are given the acceleration of the rocket and the time at which the engine gets switched off. We have to find the maximum height of the rocket. Firstly, apply the first equation of the motion to calculate the velocity of the rocket. Then, for the height at which the rocket would be reached in five seconds apply the second equation of the motion. Now, apply the third equation of motion to calculate that distance and take the initial velocity to be the final velocity of the first case and the final velocity will be zero. The total distance will be the sum of the required distances.

Formula used:

Equations of the motion –
$v = u + at$
$s = ut + \dfrac{1}{2}a{t^2}$
${v^2} = {u^2} - 2gs$ (If the acceleration is zero and gravitational force is working in the downward direction)


Complete answer:
Case 1:
Given that,
A rocket is fired upward with the acceleration ${\text{19}}{\text{.6 m}}{{\text{s}}^{{\text{ - 2}}}}$ and after ${\text{5 sec}}{\text{.}}$ engine gets switched off.
Acceleration $a = {\text{19}}{\text{.6 m}}{{\text{s}}^{{\text{ - 2}}}}$
Time $t = {\text{5 sec}}{\text{.}}$
Initial velocity $u = 0$
Using, the first equation of motion i.e., $v = u + at$
We get, $v = 0 + \left( {19.6} \right)\left( 5 \right)$
$v = 98{\text{ m}}{{\text{s}}^{ - 1}}$
Now, for the distance apply second equation of the motion i.e., $s = ut + \dfrac{1}{2}a{t^2}$
Substituting the required values,
We get, $s = \dfrac{1}{2}\left( {19.6} \right){\left( 5 \right)^2}$
$s = 245{\text{ m}}$
Case 2:
Now, let the height be $h$ when the engine get switched off and moves in upward direction. Also, the velocity of the rocket will be zero.
It implies that, initial velocity $u = 98{\text{ m}}{{\text{s}}^{ - 1}}$
Final velocity $v = 0{\text{ m}}{{\text{s}}^{ - 1}}$
Height $ = h$
Acceleration due to gravity $g = {\text{9}}{\text{.8 m}}{{\text{s}}^{{\text{ - 2}}}}$
Using third equation of the motion i.e., ${v^2} = {u^2} - 2gs$
Substitution the required values,
We get, $h = \dfrac{{{{\left( {98} \right)}^2}}}{{2\left( {9.8} \right)}}$
So, $h = 490{\text{ m}}$
Therefore, the maximum height rocket reached from the earth’s surface will be
$s + h = 245 + 490 = 735{\text{ m}}$
Hence, option (B) is the correct answer i.e., ${\text{735 m}}$.







Note:In physics, motion is the change in position or orientation of a body over time. Translation is movement along a straight line or a curve. Rotation is defined as motion that alters the orientation of a body. All places in the body have the same velocity (directed speed) and acceleration in both circumstances (time rate of change of velocity). The most common type of motion involves translation and rotation. Every movement is relative to some frame of reference. Saying that a body is at rest, or that it is not in motion, simply indicates that it is being described in relation to a frame of reference that is moving together with the body.