183_notes:examples:finding_the_range_of_projectile [Projects & Practices in Physics]

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===== 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|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 ====

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. 

$$ x_f = x_i + V_{avg,x} \Delta{t}$$

Plug in respective values for variables.

$$ = 0 + 80m/s(3.65s)$$

Compute range in x-direction.
       
$$ = 292m$$

Repeat same process for the z-components:
      
$$ z_f = z_i + V_{avg,z} \Delta{t}$$ 

Plug in respective values for variables.
    
$$ = -5 + -5m/s(3.65s)$$

Compute range in z-direction.
      
$$ = -23.25m$$ 

Write range(final position vector) using all components:     
       
Final position = $$\langle 292,0,-23.255 \rangle m $$