# A model rocket is fired vertically from rest. It has a constant acceleration of 17.5m/s^2 for the first 1.5 s. Then its fuel is exhausted, and it is in free fall. (a) Ignoring air resistance, how high does the rocket travel? Cont.

## (b) How long after liftoff does the rocket return to the ground?

Feb 11, 2017

(a) Given acceleration $a = 15.5 m {s}^{-} 2$ for first 1.5 seconds, after starting from rest. Let $v$ be velocity attained when fuel is exhausted. Kinematic equation is
$v = u + a t$ ........(1)

Inserting given values we get
$v = 0 + 17.5 \times 1.5$
$\implies v = 27.25 m {s}^{-} 1$ .....(2)

Using the following kinematic equation for finding height attained till $1.5 s$
${v}^{2} - {u}^{2} = 2 a s$
${\left(27.25\right)}^{2} - {0}^{2} = 2 \times 17.5 h$
$\implies h = {\left(27.25\right)}^{2} / \left(2 \times 17.5\right)$
$\implies h \approx 21.22 m$ ......(3)

These equations (2) and (3) give initial conditions for the freely falling rocket after fuel is exhausted.

Let rocket reach a maximum height $\left(h + {h}_{1}\right)$ where velocity is zero. Acceleration due to gravity is in a direction opposite to the positive direction of motion.
To calculate height ${h}_{1}$ attained under free fall we use the kinematic relation
${v}^{2} - {u}^{2} = 2 a s$ ........(4)
$\therefore$ ${0}^{2} - {\left(27.25\right)}^{2} = 2 \times \left(- 9.81\right) {h}_{1}$
$\implies {h}_{1} = {\left(27.25\right)}^{2} / 19.62$
$\implies {h}_{1} \approx 37.85 m$

Maximum height attained is $h + {h}_{1} = 21.22 + 37.85 = 59.07 m$

(b) Let time taken to travel from height $h$ to height ${h}_{1}$ be ${t}_{1}$.
It can be found from the kinematic equation (1)
$0 = 27.25 - 9.81 {t}_{1}$
$\implies {t}_{1} = \frac{27.25}{9.81} = 2. \overline{7} s$

Now for downward journey of rocket, let the time taken for falling from maximum height to the ground be ${t}_{2}$. Applicable kinematic expression is
$s = u t + \frac{1}{2} a {t}^{2}$ ......(4)
Acceleration due to gravity is in the direction of motion. We have
$59.07 = 0 \times t + \frac{1}{2} \left(9.81\right) {t}_{2}^{2}$
$\implies {t}_{2}^{2} = \sqrt{\frac{59.07 \times 2}{9.81}}$
$\implies {t}_{2}^{2} = \approx 3.47 s$

Total time taken after liftoff$= 1.5 + {t}_{1} + {t}_{2}$
$= 1.5 + 2. \overline{7} + 3.47 = 7.7 s$, rounded to one decimal place

-.-.-.-.-.-.-.-.-.-.-.

Alternate method for part (b)
After $1.5 s$. Rocket is falling freely under gravity. When it reaches ground height $= 0$.
Displacement$= \text{Final position" -"Initial position}$
$= 0 - 21.22 = - 21.22 m$
Time taken to reach ground can be calculated using (4). Acceleration due to gravity acting against the direction of motion.
$- 21.22 = 27.25 t + \frac{1}{2} \left(- 9.81\right) {t}^{2}$
$\implies 9.81 {t}^{2} - 54.5 t - 42.44 = 0$
Roots of this quadratic can be found using

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Using inbuilt graphic tool.

Ignoring the negative root as time can not be negative. we have
Time of flight $t$ after fuel is consumed $= 6.2 s$, rounded to one decimal place.
Total time taken after liftoff$= 1.5 + t$
$= 1.5 + 6.2 = 7.7 s$, rounded to one decimal place