Question

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2 for 2.30 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

Answer 1

Answer:

Maximum height attained by the model rocket is 2172.87 m

Explanation:

Given,

• Initial speed of the model rocket = u = 0
• acceleration of the model rocket = $a\ =\ 85.0 m/s^2$
• time during the acceleration = t = 2.30 s

We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 $m/s^2$ for the time t = 2.30 s before the fuel abruptly runs out.

Let $s_1$ be the height attained by the rocket during this time intervel,

$s_1\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s_1\ =\ 0\ +\ 0.5\times 85\times 2.30^2\\\Rightarrow s_1\ =\ 224.825\ m$

And Final velocity at that point be v

$\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.$

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let '$s_2$' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

• initial velocity of the rocket = v = 195.5 m/s.

• a = $-g\ =\ -9.81\ m/s^2$
• Final velocity of the rocket at the maximum altitude = $v_f\ =\ 0$

From the kinematics,

$v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m$

Hence the maximum altitude attained by the rocket from the ground is

$s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m$