# Question 33891

Jan 5, 2017

$\textsf{102.5 \textcolor{w h i t e}{x} s}$

#### Explanation:

I will consider the motion in two stages.

1. The Powered Stage

To get the final velocity at the end of the powered stage we can use:

$\textsf{v = u + a t}$

This becomes:

$\textsf{v = 0 + \left(\frac{9.81}{3}\right) \times 60 = 196.2 \textcolor{w h i t e}{x} \text{m/s}}$

To find the height h reached we can use:

$\textsf{{v}^{2} = {u}^{2} + 2 a s}$

This becomes:

$\textsf{{196.2}^{2} = 0 + \frac{2 \times 9.81}{3} \times h}$

$\therefore$$\textsf{h = \frac{3.849 \times {10}^{4} \times 3}{2 \times 9.81} = 588.6 \textcolor{w h i t e}{x} m}$

2. The Unpowered Stage

The rocket is now moving under the influence of gravity.

We can use:

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

I will use the convention that "up is positive".

The equation becomes:

$\textsf{- 588.6 = 196.2 t - \frac{1}{2} \times 9.81 \times {t}^{2}}$

$\therefore$$\textsf{4.9 {t}^{2} - 196.2 t - 588.6 = 0}$

sf(t=(196.2+-sqrt(37,094.76-[4xx4.9xx(-588.6)]))/(9.8)#

$\textsf{t = \frac{196.2 \pm 220.5}{9.8}}$

Ignoring the -ve root:

$\textsf{t = 42.5 \textcolor{w h i t e}{x} s}$

To get the total time of flight $\left(\textsf{{t}_{\text{tot}}}\right)$, we add this to the time taken for the powered part of the flight which we know is 1 minute.

$\therefore$$\textsf{{t}_{\text{tot}} = 60 + 42.5 = 102.5 \textcolor{w h i t e}{x} s}$