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183_notes:examples:finding_the_range_of_projectile [2014/07/20 22:17] – pwirving | 183_notes:examples:finding_the_range_of_projectile [2014/07/23 05:36] – pwirving | ||
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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 $\langle 0,40,-5 \rangle$ 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: |
=== Facts ==== | === 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: | ||
=== 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: | ||
+ | |||
+ | 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: | ||
==== Solution ==== | ==== 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 |
+ | |||
+ | We now to find the range in the x and z directions | ||
+ | |||
+ | 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. | ||
+ | |||
+ | $$ = 0 + 80m/ | ||
- | $$ V_{fv} = V_{iy} + (\dfrac{F_{net, | + | Compute range in x-direction. |
- | + | | |
- | $$ = V_{iy} + (\dfrac{-mg}{m}) \Delta{t}$$ | + | $$ = 767m$$ |
- | $$ = V_{iy} - g\Delta{t}$$ | + | Repeat same process for the z-components: |
+ | |||
+ | $$ z_f = z_i + V_{avg,z} \Delta{t}$$ | ||
- | $$ = 7m/s - (9.8 \dfrac{N}{kg})(9.59s)$$ | + | Plug in respective values for variables. |
+ | |||
+ | $$ = -5 + -5m/s(9.59s)$$ | ||
- | $$ = -87m/s$$ | + | Compute range in z-direction. |
+ | |||
+ | $$ = -52.95$$ | ||
+ | Write range(final position vector) using all components: | ||
+ | |||
+ | Final position = $$\langle 767, |