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--- 183_notes:examples:finding_the_range_of_projectile [2014/07/21 00:10] – pwirving
+++ 183_notes:examples:finding_the_range_of_projectile [2015/09/17 12:16] (current) – caballero
@@ Line 1: / Line 1: @@
 ===== Example: Finding the range of a projectile =====
-For the previous example of the out of control bus which is forced to jump from a location  $\langle 0,40,-5 \rangle$  with an initial velocity of  $\langle 80,7,-5 \rangle$ . We have now found the time of flight to be 9.59s and now want to find 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|3.65s]] and now want to find the position of where the bus returns to the ground.
 === Facts ====
-* Starting position of the bus  $\langle 0,40,-5 \rangle$
+  * Starting position of the bus  $\langle 0,40,-5 \rangle$
+  * Initial velocity of the bus  $\langle 80,7,-5 \rangle$
-* Initial velocity of the bus  $\langle 80,7,-5 \rangle$
+  * 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 acceleration due to gravity is 9.8  $\dfrac{m}{s^2}$  and is directed downward.
+  * The bus takes [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|3.65s]] to reach the ground (from previous problem)
-* The bus experiences one force - the gravitational force (directly down).
-* The bus takes 9.59s to reach the ground
 === Lacking ===
-* The final position of the bus.
+  * The final position of the bus.
 === Approximations & Assumptions ===
-* Assume no drag effects
+  * Assume no drag effects
+  * Assume ground is when position of bus is 0 in the y direction
-* Assume ground is when position of bus is 0 in the y direction
 === Representations ===
-Diagram of situation.
-{{183_notes:bus_vectors_2nd_part.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 ====
@@ Line 43: / Line 36: @@
 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.
  $x_f = x_i + V_{avg,x} \Delta{t}$
-$$ = 0 + 80m/s(9.59s)$$
+Plug in respective values for variables.
+$$ = 0 + 80m/s(3.65s)$$
+Compute range in x-direction.
-$$ = 767m$$
+$$ = 292m$$
+Repeat same process for the z-components:
  $z_f = z_i + V_{avg,z} \Delta{t}$
+Plug in respective values for variables.
-$$ = -5 + -5m/s(9.59s)$$
+$$ = -5 + -5m/s(3.65s)$$
+Compute range in z-direction.
-$$ = -52.95$$
+$$ = -23.25m$$
-Final position =  $\langle 767,0,-52.95 \rangle$  m
+Write range(final position vector) using all components:
+Final position =  $\langle 292,0,-23.255 \rangle m$