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| Both sides previous revision Previous revision Next revision | Previous revisionLast revisionBoth sides next revision | ||
| 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2014/07/22 06:12] – pwirving | 183_notes:examples:finding_the_time_of_flight_of_a_projectile [2015/09/15 16:46] – [Solution] pwirving | ||
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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: | + | {{183_notes: |
| General equation for position up date formula for constant force systems: | General equation for position up date formula for constant force systems: | ||
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| $$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$ | $$\vec{r}_{f} = \vec{r}_{i} + \vec{v}_{i} \Delta t + \dfrac{1}{2}\dfrac{\vec{F}_{net}}{m} \Delta t^2$$ | ||
| - | Equation for determining $\Delta{t}$ in constant force situations.<wrap todo>use vector principle; then motivate why only y matters</ | + | Equation for determining $\Delta{t}$ in constant force situations. |
| ==== Solution ==== | ==== Solution ==== | ||
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| The bus has an initial velocity in the x-direction of $80m/ | The bus has an initial velocity in the x-direction of $80m/ | ||
| - | This x-velocity has no effect on the amount of time it takes for the bus to reach the ground. | + | This x-velocity has no effect on the amount of time it takes for the bus to reach the ground. Just as the z-velocity does not either. |
| The force due to gravity is constant so we can use the equation for position up date formula for constant force systems. | The force due to gravity is constant so we can use the equation for position up date formula for constant force systems. | ||
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| At ground $y_f = 0$; the bus initially leaves the road at $y_i = 40$. | At ground $y_f = 0$; the bus initially leaves the road at $y_i = 40$. | ||
| - | |||
| - | $y_f - y_i = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{F_{net, | ||
| $0 - 40 = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{-mg}{m} \Delta{t^2}$ | $0 - 40 = v_{iy} \Delta{t} + \dfrac{1}{2} \dfrac{-mg}{m} \Delta{t^2}$ | ||
| + | |||
| + | The masses cancel. | ||
| $ -40 = v_{iy} \Delta{t} + -\dfrac{1}{2} {g} \Delta{t^2}$ | $ -40 = v_{iy} \Delta{t} + -\dfrac{1}{2} {g} \Delta{t^2}$ | ||
| + | |||
| + | Rearrange the equation. | ||
| $ -40 -v_{iy} \Delta{t} = -\dfrac{1}{2} {g} \Delta{t^2}$ | $ -40 -v_{iy} \Delta{t} = -\dfrac{1}{2} {g} \Delta{t^2}$ | ||
| - | $ -40 -v_{iy} | + | Sub in values for $v_{iy}$ and ${g}$ |
| + | |||
| + | $ -40 -(7 \Delta{t}) = -4.9 \Delta{t^2}$ | ||
| + | |||
| + | Substitute in value for $v_{iy}$ and g | ||
| $ 2\dfrac{{40 + v_{iy}}}{g} = \Delta{t}$ | $ 2\dfrac{{40 + v_{iy}}}{g} = \Delta{t}$ | ||
| - | $ \Delta{t} = 9.59s$ | + | Compute time it takes for the bus to reach the ground. |
| + | $ \Delta{t} = 9.59s$ | ||