From 787067e79dd163b9313011a8af52061a065166ad Mon Sep 17 00:00:00 2001 From: kb Date: Mon, 17 Jul 2023 20:15:53 +1000 Subject: [PATCH] CM1015 midterm WIP --- notes/study/CM1015/CM1015-midterm.md | 34 +++++++++++++++++++++++++++- 1 file changed, 33 insertions(+), 1 deletion(-) diff --git a/notes/study/CM1015/CM1015-midterm.md b/notes/study/CM1015/CM1015-midterm.md index 70dd0aa..c52aaaf 100644 --- a/notes/study/CM1015/CM1015-midterm.md +++ b/notes/study/CM1015/CM1015-midterm.md @@ -994,7 +994,39 @@ We can then visualise the range of $g(t)$ spans $[1,9]$. ## Question 5 (c) -==TODO - kinematics== +To discover the maximum height reached by the banana box it is assumed this scenario exists with the renowned ELOBSA (Extreme Low Orbit Banana Satellite Association), where the only force acting on it is gravity at -9.807m/s/s. + +Although it will transit the perfectly fiction-less curved test surface of their advanced vacuum chamber, any lateral movement becomes unimportant due to the isolation from all other affecting energy sources. + +Using the kinematic formula that relates change in velocity, acceleration, and displacement we may solve for the point of zero velocity where the vehicle under test will have reached the peak of the flight test. + +$$+v^{2} = u^{2} + 2 \cdot a \cdot d +$$ + +where ++\begin{align*} +a &: \text{acceleration (const.)}\\ +u &: \text{initial velocity}\\ +v &: \text{final velocity}\\ +d &: \text{displacement} +\end{align*} + + ++\begin{align*} +0^{2} &= 40^{2}+ 2 \cdot -9.807 \cdot d\\ +\\ +0 &= 1602 - 19.614d\\ +\\ +d &= \frac{1602}{19.614}\\ +\\ +&\approx 81.676 +\end{align*} + + +Under the conditions stated, with a nominal launch we may expect the test vehicle to reach 81.876 meters (approximately 459 bananas for scale) of elevation on the track before beginning an uncontrolled descent. ## Question 5 (d) -- 2.43.0