183_notes:examples:finding_the_range_of_projectile

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183_notes:examples:finding_the_range_of_projectile [2014/07/20 23:01] – 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 =====
  
-For the previous example of the out of control bus which is forced to jump from a location ⟨0,40,−5⟩ with an initial velocity of ⟨80,7,−5⟩. 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 ⟨0,40,−5⟩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 ====
 +
 +  * Starting position of the bus ⟨0,40,−5⟩
 +  * Initial velocity of the bus ⟨80,7,−5⟩
 +  * The acceleration due to gravity is 9.8 ms2 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 === === Lacking ===
 +
 +  * The final position of the bus.
  
 === Approximations & Assumptions === === Approximations & Assumptions ===
 +
 +  * Assume no drag effects
 +  * Assume ground is when position of bus is 0 in the y direction
  
 === Representations === === 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:
 +
 +→rf=→ri+→vavgΔt
 +
 +Also know as the [[183_notes:displacement_and_velocity|position update formula]]. 
  
 ==== Solution ==== ==== Solution ====
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 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. 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. 
  
 xf=xi+Vavg,xΔt xf=xi+Vavg,xΔt
  
-  $$ = 0 + 80m/s(9.59s)$$+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:
              
 zf=zi+Vavg,zΔt  zf=zi+Vavg,zΔt 
 +
 +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 = ⟨767,0,−52.95⟩ m+
  
 +Write range(final position vector) using all components:     
 +       
 +Final position = ⟨292,0,−23.255⟩m 
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