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--- 183_notes:examples:finding_the_range_of_projectile [2014/07/22 06:22] – pwirving
+++ 183_notes:examples:finding_the_range_of_projectile [2014/07/23 05:36] – pwirving
@@ Line 1: / Line 1: @@
 ===== Example: Finding the range of a projectile =====
-In the previous example of [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|time of flight]], the out of control bus is forced to jump from a location $\langle 0,40,-5 \rangle$m with an initial velocity of $\langle 80,7,-5 \rangle m/s^-1$. We have now found the time of flight to be [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|9.59s]] and now want to find the position of where the bus returns to the ground.
+In the previous example of [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|time of flight]], the out of control bus is forced to jump from a location $\langle 0,40,-5 \rangle$m with an initial velocity of $\langle 80,7,-5 \rangle m/s^{-1}$. We have now found the time of flight to be [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|9.59s]] and now want to find the position of where the bus returns to the ground.
 === Facts ====
@@ Line 9: / Line 9: @@
   * 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).
-  * The bus takes 9.59s to reach the ground (from previous problem <wrap todo>link</wrap>)
+  * The bus takes [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|9.59s]] to reach the ground (from previous problem)
 === Lacking ===
@@ Line 21: / Line 21: @@
 === Representations ===
-Diagram of situation.<wrap todo>confused by what's labeled</wrap>
-{{183_notes:examples:bus2.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 calculating the final position of an object.
+The general equation for calculating the final position of an object:
-$$ x_f = x_i + V_{avg,x} \Delta{t}$$
+$$ \vec{r}_f = \vec{r}_i + \vec{v}_{avg} \Delta t $$
+Also know as the [[183_notes:displacement_and_velocity|position update formula]].
 ==== Solution ====
-<WRAP todo> A little more commentary on the problem, which equations are you using and why?</WRAP>
 From the previous problem you already know the final location of the ball in the y direction to be 0 as it has met the ground after 9.59s.
-Now to find the range in the x and z directions:
+We now to find the range in the x and z directions in order to have a position vector for the final resting place of the bus.
+There is no force acting in the x or z directions as the only force acting on the system is the gravitational force which acts in the y-direction.
+This means that the initial velocities in both of these directions have remained unchanged.
+We know the amount of time the bus has been traveling in the x-direction at its initial velocity and its initial position so we can compute the distance travelled in this direction using the position update formula for x-components.
 $$ x_f = x_i + V_{avg,x} \Delta{t}$$
+Plug in respective values for variables.
 $$ = 0 + 80m/s(9.59s)$$
+Compute range in x-direction.
 $$ = 767m$$
+Repeat same process for the z-components:
 $$ z_f = z_i + V_{avg,z} \Delta{t}$$
+Plug in respective values for variables.
 $$ = -5 + -5m/s(9.59s)$$
+Compute range in z-direction.
 $$ = -52.95$$
+Write range(final position vector) using all components:
-Final position = $\langle 767,0,-52.95 \rangle$ m
+Final position = $$\langle 767,0,-52.95 \rangle m $$