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--- 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2014/07/22 06:13] – pwirving
+++ 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2015/09/15 16:46] – [Solution] pwirving
@@ Line 24: / Line 24: @@
 Diagram of forces acting on bus once it leaves the road.
-{{183_notes:bus_force.jpg}}
+{{183_notes:examples:bus_abstract.jpg}}
 General equation for position up date formula for constant force systems:
@@ Line 30: / Line 30: @@
 $$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$
-Equation for determining $\Delta{t}$ in constant force situations.<wrap todo>use vector principle; then motivate why only y matters</wrap>
+Equation for determining $\Delta{t}$ in constant force situations.
 ==== Solution ====
@@ Line 51: / Line 51: @@
 At ground $y_f = 0$; the bus initially leaves the road at $y_i = 40$.
-$y_f - y_i = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{F_{net,y}}{m} \Delta{t^2}$
 $0 - 40 = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{-mg}{m} \Delta{t^2}$
+The masses cancel.
 $ -40 = v_{iy} \Delta{t} + -\dfrac{1}{2} {g} \Delta{t^2}$
+Rearrange the equation.
 $ -40 -v_{iy} \Delta{t} = -\dfrac{1}{2} {g} \Delta{t^2}$
-$ -40 -v_{iy} = -\dfrac{1}{2} {g} \Delta{t}$
+Sub in values for $v_{iy}$ and ${g}$
+$ -40 -(7 \Delta{t}) = -4.9 \Delta{t}$
+Substitute in value for $v_{iy}$ and g
 $ 2\dfrac{{40 + v_{iy}}}{g} = \Delta{t}$
-$ \Delta{t} = 9.59s$
+Compute time it takes for the bus to reach the ground.
+$ \Delta{t} = 9.59s$