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--- 183_notes:examples:finding_the_range_of_projectile [2014/07/20 22:24] – 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  $\langle 0,40,-5 \rangle$
+  * 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 bus takes [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|3.65s]] to reach the ground (from previous problem)
 === Lacking ===
+  * The final position of the bus.
 === Approximations & Assumptions ===
+  * Assume no drag effects
+  * Assume ground is when position of bus is 0 in the y direction
 === Representations ===
+Diagram of forces acting on bus once it leaves the road.
+{{183_notes:examples:bus_abstract.jpg}}
+The general equation for calculating the final position of an object:
+ $\vec{r}_f = \vec{r}_i + \vec{v}_{avg} \Delta t$
+Also know as the [[183_notes:displacement_and_velocity|position update formula]].
 ==== Solution ====
-First find  $v_{fy}$  for when it hits the ground. We need this in order to find  $\vec{v_{avg}}$
+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.
+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.
-$$ V_{fv} = V_{iy} + (\dfrac{F_{net,y}}{m}) \Delta{t}$$
+$$ x_f = x_i + V_{avg,x} \Delta{t}$$
-$$        = V_{iy} + (\dfrac{-mg}{m}) \Delta{t}$$
- $= V_{iy} - g\Delta{t}$
+Plug in respective values for variables.
-$$        = 7m/s - (9.8 \dfrac{N}{kg})(9.59s)$$
+$$ = 0 + 80m/s(3.65s)$$
-$$        = -87m/s$$
+Compute range in x-direction.
+$$ = 292m$$
-Now to find the range:
+Repeat same process for the z-components:
+ $z_f = z_i + V_{avg,z} \Delta{t}$
-$$ \vec{r_f} = \vec{r_i} + \vec{v_{avg}}\Delta{t}$$
+Plug in respective values for variables.
+$$ = -5 + -5m/s(3.65s)$$
-$$           = \vec{r_i} + \dfrac{\vec{v_i} + \vec{v_f}}{2} \Delta{t}$$
+Compute range in z-direction.
+$$ = -23.25m$$
+Write range(final position vector) using all components:
+Final position =  $\langle 292,0,-23.255 \rangle m$