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--- 183_notes:examples:mit_water_balloon_fight [2014/10/10 20:29] – pwirving
+++ 183_notes:examples:mit_water_balloon_fight [2014/10/11 06:03] (current) – pwirving
@@ Line 39: / Line 39: @@
 Surroundings: Surgical hose slingshot
-{{course_planning:|}}
+{{course_planning:projects:water_balloon_fight.jpg?300}}
 === Solution ===
@@ Line 81: / Line 81: @@
 And the initial and final conditions are:
-Initial  = S_{max} \longrightarrow max stretch
+Initial  = $S_{max} \longrightarrow$ max stretch
-Final = 0 \longrightarrow relaxed
+Final = $0 \longrightarrow$ relaxed
 Sub in the force for hose into the equation for work done by a non constant force:
@@ Line 97: / Line 97: @@
 Going back a couple of steps we now have an equation for work which we can relate to our previous equation for v_{f}.
-$$V_{f} = \sqrt {\dfrac{2W}{m}}$$ ; $$W = \dfrac{1}{2}S_{max}^2$$
+$$V_{f} = \sqrt {\dfrac{2W}{m}}$$
+$$W = \dfrac{1}{2}S_{max}^2$$
+Substitute work in the equation for $v_{f}$
 $$V_{f} = \sqrt {\dfrac{2}{m}(\dfrac{1}{2}ks^2_{max})}$$
@@ Line 105: / Line 109: @@
 $$ = \sqrt {\dfrac{k}{m}}S_{max}$$
-Sub in values for known variables and solve for v_{f}
+Sub in values for known variables and solve for $v_{f}$
 $$ v_{f} = \sqrt {\dfrac{100N/m}{0.5kg}}  (3.5m)^2 \longrightarrow 50m/s = 111mph$$