Question

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After this point, its acceleration is that of gravity, downward. (a) What is the velocity of the rocket when it runs out of fuel? (Assume up is positive.)
1 m/s
(b) How long does it take to reach this point?
2 s
(c) What maximum altitude does the rocket reach?
3 m
(d) How much time (total) does it take to reach maximum altitude?
4 s
(e) With what velocity does the rocket strike the Earth? (Assume up is positive.)
5 m/s
(f) How long (total) is it in the air?
6 s

Answer 1

Concepts and reason

The concept required to solve this question is velocity, distance and time.

Initially, if the object is starting from rest then the initial speed of the object is equal to zero. Later, from the linear kinematic expression calculate the velocity of the rocket, maximum altitude and finally, the time took to reach the maximum altitude.

Fundamentals

From the linear kinematic expression the expression of distance covered by the object in terms of the acceleration, initial velocity and time is equal to,

$s = ut + \frac{1}{2}a{t^2}$

Here,$s$is the distance,$u$is the initial speed, $a$is the acceleration, and $t$is the time.

The expression of the final velocity in terms of the initial speed of the object is expressed as follows,

$v = u + at$

Here,$v$is the final velocity,$u$is the initial velocity and, $t$is the time.

The expression of the final velocity in terms of the distance is equal to,

${v^2} = {u^2} + 2as$

(a)

The expression of the velocity when the rocket runs out of fuel is equal to,

${v^2} = {u^2} + 2a\left( {{y_1} - {y_0}} \right)$

Substitute $0.00{\rm{ m/s}}$for$u$,$3.2{\rm{ m/}}{{\rm{s}}^2}$for$a$, and $1300{\rm{ m}}$for${y_1} - {y_0}$in the above expression of the final velocity of the rocket.

$\begin{array}{c}\\{v^2} = 0 + \left( 2 \right)\left( {3.2{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1300{\rm{ m}}} \right)\\\\ = 8320{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\\end{array}$

Taking the square root on both sides,

$\begin{array}{c}\\v = \sqrt {8320{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\\\ = 91.21{\rm{ m/s}}\\\end{array}$

[Part a]

Part a

(b)

The expression of the time taken to reach the maximum altitude is equal to,

$v = u + a{t_1}$

Rearrange the above expression of the final velocity in terms of the time, that is,

${t_1} = \frac{{v - u}}{a}$

Substitute $91.21{\rm{ m/s}}$for$v$,$0$for$u$, and $3.2{\rm{ m/}}{{\rm{s}}^2}$for$a$in the above expression of the time taken.

$\begin{array}{c}\\{t_1} = \frac{{91.21{\rm{ m/s}} - 0}}{{3.2{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 28.5{\rm{ s}}\\\end{array}$

(c)

The expression of the final velocity in when acceleration due to the gravity is downward.

${v^2} = {u^2} - 2g\left( {h - \left( {{y_1} - {y_0}} \right)} \right)$

Substitute $0.00{\rm{ m/s}}$ for $v$,$91.21{\rm{ m/s}}$ for $u$,$9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$, and $1300{\rm{ m}}$ for $\left( {{y_1} - {y_0}} \right)$ in the above expression of the final velocity expression.

$0 = {\left( {91.21{\rm{ m/s}}} \right)^2} - \left( 2 \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {h - 1300{\rm{ m}}} \right)$

Rearrange the above expression in term of the height$h$.

$\begin{array}{c}\\h = 1300{\rm{ m}} + \frac{{{{\left( {91.21{\rm{ m/s}}} \right)}^2}}}{{\left( 2 \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)}}\\\\ = 1724{\rm{ m}}\\\end{array}$

(d)

The expression of the final velocity in terms of the acceleration and time is equal to,

$v = u - g\left( {t - {t_1}} \right)$

Substitute $91.21{\rm{ m/s}}$ for $u$,$0$for$v$,$9.8{\rm{ m/}}{{\rm{s}}^2}$, and $28.5{\rm{ s}}$ for ${t_1}$in the above expression of the time taken to reach the maximum altitude.

$0 = 91.21{\rm{ m/s}} - \left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {t - 28.5{\rm{ s}}} \right)$

Rearrange the above expression in terms of time$t$.

$\begin{array}{c}\\t = 28.5{\rm{ s}} + \frac{{91.21{\rm{ m/s}}}}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 37.8{\rm{ s}}\\\end{array}$

(e)

The expression for the final velocity in terms of the acceleration due to the gravity.

${v^2} = {u^2} + 2g\left( {{y_0} - {y_1}} \right)$

Substitute $91.21{\rm{ m/s}}$ for $u$,$9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$, and $1300{\rm{ m}}$ for ${y_0} - {y_1}$ in the above expression of the final velocity.

$\begin{array}{c}\\{v^2} = {\left( {91.21{\rm{ m/s}}} \right)^2} + \left( 2 \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {1300{\rm{ m}}} \right)\\\\ = 33800{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\\end{array}$

Taking the square root on both sides,

$\begin{array}{c}\\v = \sqrt {33800{\rm{ }}{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\\\ = 183.8{\rm{ m/s}}\\\end{array}$

(f)

The expression of the time in which rocket is in air is expressed as follows,

$v = u + g\left( {{t_3} - {t_1}} \right)$

Substitute $- 183.5{\rm{ m/s}}$ for $v$,$- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$, $91.21{\rm{ m/s}}$for $u$, and $28.5{\rm{ s}}$ for $t$ in the above expression of the final velocity.

$- 183.5{\rm{ m/s}} = 91.21{\rm{ m/s}} + \left( { - 9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\left( {{t_3} - 28.5{\rm{ s}}} \right)$

Rearrange the above expression in terms of the time${t_3}$.

$\begin{array}{c}\\{t_3} = 28.5{\rm{ s}} + \frac{{91.21{\rm{ m/s}} + 183.5{\rm{ m/s}}}}{{9.8{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 56.5{\rm{ s}}\\\end{array}$

Ans: Part a

The velocity of the rocket when the rocket runs out of fuel is equal to$91.21{\rm{ m/s}}$.