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--- 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2014/07/22 05:45] – pwirving
+++ 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2015/09/15 16:46] – [Solution] pwirving
@@ Line 5: / Line 5: @@
 === Facts ====
-  * Starting position of the bus $\langle 0,40,-5 \rangle$
+  * Starting position of the bus $\langle 0,40,-5 \rangle$m
-  * Initial velocity of the bus $\langle 80,7,-5 \rangle$
+  * Initial velocity of the bus $\langle 80,7,-5 \rangle$m
   * The acceleration due to gravity is 9.8 $\dfrac{m}{s^2}$ and is directed downward.
   * The bus experiences one force - the gravitational force (directly down).
@@ Line 21: / Line 21: @@
 === Representations ===
-Diagram of the situation.<wrap todo> I'm confused with what is labeled here</wrap>
-{{183_notes:examples:bus_vectors.jpg}}
 Diagram of forces acting on bus once it leaves the road.
-{{183_notes:bus_force.jpg}}
+{{183_notes:examples:bus_abstract.jpg}}
-Equation for determining $\Delta{t}$ in constant force situations.<wrap todo>use vector principle; then motivate why only y matters</wrap>
+General equation for position up date formula for constant force systems:
-$y_f - y_i = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{F_{net,y}}{m} \Delta{t^2}$
+$$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$
+Equation for determining $\Delta{t}$ in constant force situations.
 ==== Solution ====
-At ground $y_f = 0$; the bus initially leaves the road at $y_i = 40$.
+In this problem we know the only force acting on the bus is the force due to gravity which acts solely in the y-direction.
+The bus has an initial velocity in the x-direction of $80m/s^{-1}$ but with no forces acting in the x-direction this velocity will remain constant.
+This x-velocity has no effect on the amount of time it takes for the bus to reach the ground. Just as the z-velocity does not either.
+The force due to gravity is constant so we can use the equation for position up date formula for constant force systems.
+We know the final position of the bus is the ground where the y-component of the position vector is 0.
+We know the bus is accelerating towards the ground at $-9.8ms^{-2}$ due to the net force in the y-direction being the acceleration due to the force of gravity.
+Since we are concerned only with the y-component of the motion we can use the y-component version of the position update equation:
 $y_f - y_i = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{F_{net,y}}{m} \Delta{t^2}$
+At ground $y_f = 0$; the bus initially leaves the road at $y_i = 40$.
 $0 - 40 = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{-mg}{m} \Delta{t^2}$
+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^2}$
+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$