# Question #754ff

Sep 14, 2016

Given
$u \to \text{Initial velocity of rocket} = 0 \frac{m}{s}$

$a \to \text{Acceleration of the rocket} = 2.10 m {s}^{-} 2$
$t \to \text{Duration of upward acceleration} = 30 s$

Velocity gained in 30s is

$v = u + a \times t = 0 + 2.1 \cdot 30 = 63 \frac{m}{s}$

1.The engine fails at this moment and rocket undergoes free fall under gravity.If it reaches height h at this point then

$h = u \times t + \frac{1}{2} \times a \times {t}^{2}$
$= 0 \cdot 30 + 0.5 \cdot 2.1 \cdot {30}^{2} = 945 m$

After 30s of its journey it will go up with retardation $g = - 9.8 m {s}^{-} 2$ till its velocity becomes zero.

2.At this highest point it will have a downward acceleration $g = 9.8 m {s}^{-} 2$ and vlocity$= 63 \frac{m}{s}$
If it ascends a height $h '$during last phase of its ascent then
${0}^{2} = {63}^{2} - 2 \cdot 9.8 \cdot h '$
$h ' = {63}^{2} / \left(2 \cdot 9.8\right) = 202.5 m$

If the rocket takes Ts to fall back at launch pad after its engine fails then

Displacement during Ts is $= - h = - 945 m$
Initial velocity upward $= + 63 \frac{m}{s}$
And $g = - 9.8 m {s}^{-} 2$

So

$- 945 = 63 \cdot T - \frac{1}{2} \cdot 9.8 \cdot {T}^{2}$

$\implies 4.9 {T}^{2} - 63 T - 945 = 0$

$\implies T = \frac{63 + \sqrt{{63}^{2} + 4 \cdot 4.9 \cdot 945}}{2 \cdot 4.9}$
$\implies T = 21.73 s$

So the total time to return back at the launch pad after it was launched will be

3.$\text{Total time} = T + 30 = 21.73 + 30 = 51.73 s$

Sep 14, 2016

1.
Given Initial velocity of rocket $u = 0 m {s}^{-} 1$
Upwards Acceleration of the rocket $a = 2.10 m {s}^{-} 2$
Duration of upward acceleration $t = 30 s$

Velocity $v$ at the end of $30 s$, assuming upwards direction is positive.
$v = u + a t$
$\implies v = 0 + 2.10 \times 30 = 63 m {s}^{-} 1$

After engine fails the rocket undergoes free fall under gravity with initial velocity calculated above. Let it reach height $h$ when engine fails. Using the kinematic equation
$h = u t + \frac{1}{2} a {t}^{2}$
$= 0 \times 30 + \frac{1}{2} \times 2.10 \times {30}^{2} = 945 m$
2. Thereafter, first it goes up and is retarded under gravity $g = - 9.81 m {s}^{-} 2$ till its velocity becomes zero. Then there is free fall under gravity.
$\therefore$At its highest point it has velocity$= 0$ and acceleration $= - 9.81 m {s}^{-} 2$
3. Suppose the rocket takes ${t}_{f}$ to fall back at launch pad after its engine failed, we have

Displacement during time ${t}_{f}$
$= \text{Final position"-"Initial position} = 0 - 945 = - 945 m$
Setting up the kinematic equation we have
$s = u t + \frac{1}{2} a {t}^{2}$
$\implies - 945 = 63 \times {t}_{f} + \frac{1}{2} \times \left(- 9.81\right) \times {t}_{f}^{2}$

$\implies 4.905 {t}_{f}^{2} - 63 {t}_{f} - 945 = 0$
${t}_{f} = \frac{63 \pm \sqrt{{63}^{2} + 4 \times 4.905 \times 945}}{2 \times 4.905}$
Ignoring the $- v e$ root as time can not be negative
$\implies {t}_{f} = 21.7 s$, rounded to one decimal place
$\therefore$ the total time to return to launch pad after launch
$= {t}_{f} + 30 = 21.7 + 30 = 51.7 s$