183_notes:examples:finding_the_range_of_projectile

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183_notes:examples:finding_the_range_of_projectile [2014/07/22 06:27] – pwirving183_notes:examples:finding_the_range_of_projectile [2015/09/17 12:16] (current) – caballero
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 ===== Example: Finding the range of a projectile ===== ===== 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|3.65s]] and now want to find the position of where the bus returns to the ground. 
  
 === Facts ==== === Facts ====
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   * The acceleration due to gravity is 9.8 $\dfrac{m}{s^2}$ and is directed downward.   * 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 experiences one force - the gravitational force (directly down).
-  * The bus takes [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|9.59s]] to reach the ground (from previous problem)+  * The bus takes [[183_notes:examples:finding_the_time_of_flight_of_a_projectile|3.65s]] to reach the ground (from previous problem)
  
 === Lacking === === Lacking ===
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 Diagram of forces acting on bus once it leaves the road. Diagram of forces acting on bus once it leaves the road.
  
-{{183_notes:bus_force.jpg}}+{{183_notes:examples:bus_abstract.jpg}}
  
 The general equation for calculating the final position of an object: The general equation for calculating the final position of an object:
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 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. 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. 
  
 $$ x_f = x_i + V_{avg,x} \Delta{t}$$ $$ x_f = x_i + V_{avg,x} \Delta{t}$$
  
-$$ = 0 + 80m/s(9.59s)$$+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:
              
 $$ z_f = z_i + V_{avg,z} \Delta{t}$$  $$ z_f = z_i + V_{avg,z} \Delta{t}$$ 
 +
 +Plug in respective values for variables.
          
-$$ = -5 + -5m/s(9.59s)$$+$$ = -5 + -5m/s(3.65s)$$ 
 + 
 +Compute range in z-direction.
              
-$$ = -52.95$$      +$$ = -23.25m$$  
 + 
 +Write range(final position vector) using all components:     
                
-Final position = $\langle 767,0,-52.95 \rangle$ m+Final position = $$\langle 292,0,-23.255 \rangle m $$ 
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